LIBRARY 


UNIVERSITY  OF  CALIFORNIA. 


Class 


THE   ELEMENTS 

OF   THE 

MECHANICS    OF   MATERIALS 

AND    OF 

POWER  TRANSMISSION 


WILLIAM    R.    KING,    U.  S.  N.,  RETIRED 

PRINCIPAL,    BALTIMORE    POLYTECHNIC    INSTITUTE 


FIRST  EDITION 

FIRST    THOUSAND 


NEW  YORK 
JOHN  WILEY  &  SONS 

43  AND  45   EAST  igTH  ST. 

LONDON:   CHAPMAN   &   HALL,   LIMITED 

1911 


COPYRIGHT,  1911 

BY 

WILLIAM    R.    KING 


Stanhope  ipresa 

F.   H.  GILSON     COMPANY 
BOSTON.     U.S.A. 


PREFACE 

THIS  book  is  the  result  of  an  experience  of  some  years  in  class- 
room work  with  engineering  students,  and  is  intended  for  use  in 
technical  schools  and  colleges. 

It  has  frequently  occurred  to  me  that  there  is  needless  ob- 
scurity of  statement  in  the  average  engineering  text,  with  the 
consequent  discouragement  and  retardation  of  young  students. 
It  has  been  my  aim,  therefore,  to  characterize  the  demonstra- 
tions in  this  text  by  completeness  and  simplicity  of  statement, 
in  the  belief  that  such  treatment  will  greatly  facilitate  the  study 
of  more  advanced  works. 

The  Calculus  has  been  introduced  necessarily,  but  only  in  its 
elementary  form,  and  chiefly  in  demonstrations  and  solutions. 
Such  use  of  it  will  be,  I  believe,  beneficial  to  the  young  student 
in  showing  him  possibilities  in  the  application  of  the  subject  to 
practical  problems. 

The  book  is  in  two  parts.  Part  I  is  devoted  to  the  elements 
of  the  Mechanics  of  Materials,  and  Part  II  to  the  elements  of 
Power  Transmission.  It  has  been  the  aim  to  present  both 
subjects  only  to  the  extent  that  will  impart  such  a  working 
knowledge  of  the  fundamentals  as  will  enable  the  student  to 
grasp  more  extended  works  without  aid. 

Much  of  the  matter  on  iron  and  steel  in  Chapter  IX,  Part  I, 
has  been  taken  by  permission  from  Durand's  "  Practical  Marine 
Engineering,"  and  the  chief  sources  of  reference  have  been  the 
works  of  Goodman  and  Unwin. 

I  am  under  obligation  to  my  assistants  in  engineering,  William 
L.  De  Baufre,  Charles  E.  Conway,  and  Samuel  P.  Platt,  —  to 
the  first  two  for  valuable  aid  and  suggestion  and  to  the  third 
for  the  care  with  which  he  made  the  tracings  for  the  cuts. 

WILLIAM  R.   KING. 

BALTIMORE,  July  4,  1911. 

iii 


227949 


CONTENTS 

PART   I 

CHAP.  PAGE 
I.   MOMENTS.     CENTER  OF  GRAVITY.    MOMENT  OF  INERTIA.    RADIUS 

OF  GYRATION / i 

II.   BENDING  MOMENT.     BENDING-MOMENT  DIAGRAM.     SHEAR.     SHEAR 

DIAGRAM 25 

III.   THE  THEORY  OF  BEAMS.     BEAM  DESIGN  62 

IV.   THE  DEFLECTION   OF  BEAMS 73 

V.   COLUMNS.     SHAFTS 89 

VI.   INTERNAL    WORK    DUE    TO    DEFORMATION.     SUDDENLY    APPLIED 

LOADS 105 

VII.   GRAPHIC  STATICS:   SYSTEM  OF  LETTERING.     FORCE  DIAGRAM.     FU- 
NICULAR  POLYGON 114 

VIII.   FRAMED   STRUCTURES.    RECIPROCAL  DIAGRAM 145 

IX.   ENGINEERING  MATERIALS 184 

X.  TESTING  MATERIALS 207 

PART    II 

I.   TRANSMISSION  OF  POWER  BY  BELTS  AND  ROPES 219 

II.  TRANSMISSION  BY  TOOTHED  WHEELS 241 

INDEX 257* 


PART    I 

THE  ELEMENTS  OF  THE  MECHANICS 
OF    MATERIALS 


CHAPTER  I 

MOMENTS.     CENTER   OF  GRAVITY.     MOMENT    OF 
INERTIA.     RADIUS    OF   GYRATION 

1.  Introductory.  —  The   mechanics   of   materials,    embracing 
the  strength  of  materials,  is  an  all-important  subject  to  the 
engineering  student.     It  includes  all  the  calculations  connected 
with  the  design  of  machines  which  admit  of  motion  between 
some  of  their  parts  in  the  transmission  of  force,  thus  involv- 
ing   dynamical    principles ;    and   of    the   design   of    structures 
which   remain   in   the    static    state    of    rest.      It   presupposes 
for    the    student    a    course    in   mechanics,    but  the   questions 
of  moments,  center  of  gravity,  moment  of  inertia,  and  radius 
of  gyration   are  of    such    frequent  application    in    mechanical 
design  that  a  partial  review  of  those  subjects  is  given  in  this 
chapter. 

2.  Moments.  —  The  moment  of  a  force  acting  on  a  body 
may  be  defined  as  the  tendency  of  the  force  to  turn  the  body 
about  a  point,  or  about  a  fixed  axis,  and  its  measure  is  the 
product  of  the  force  by  the  perpendicular  distance  from  the 
point,   or  from   the  axis,  to   the  line   of   action  of  the  force. 
The  point,  or  axis,   about   which  the  moments  are   taken  is 
called  the  center  of  moments,  and   the  perpendicular   distance 


2  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

from  the  line  of  action  of  the  force  to  the  center  of  moments 
is  called  the  arm  of  the  force. 

If  there  be  a  number  of  forces  acting  on  the  body,  those 
tending  to  turn  it  in  one  direction  may  be  regarded  as  positive 
and  those  tending  to  turn  it  in  the  opposite  direction  as  nega- 
tive. By  common  consent  forces  with  a  turning  tendency  in  a 
clockwise  direction  are  termed  positive  and  those  with  a  contra- 
clockwise  tendency  are  termed  negative.  It  is  immaterial  which 
kind  of  force  is  termed  positive  and  which  negative,  but  having 
chosen  one  kind  as  positive  in  any  investigation  the  choice 
must  be  adhered  to  and  the  opposite  kind  must  be  regarded 
as  negative. 

In  Fig.  i  let  the  forces  P,  Q,  and  R  act  on  the  body  in  the 
directions  indicated.  If  the  body  remains  in  equilibrium  the 
underlying  principle  of  moments  asserts  that  the 
algebraic  sum  of  the  moments  of  the  forces  about 
any  point  as  a  center,  or  about  any  line  as  an 
axis,  must  be  zero,  the  point  and  the  line  being 


in  the  same  plane.     In  other  words,  the  sum  of 
the  clockwise  moments  must  equal  the  sum  of 

Fig.  i. 

the  contraclockwise  moments. 

Let  moments  be  taken  about  the  point  0.  The  force  Q  tends 
to  turn  the  body  in  a  clockwise  direction  about  0  and  will  be 
regarded  as  positive,  and  the  tendencies  of  the  forces  P  and  R 
are  to  turn  it  in  a  contraclockwise  direction  about  0  and  will 
be  regarded  as  negative.  The  equation  of  moments  will  then 

be 

Qq  -  Pp  -  Rr  =  o. 

This  follows  directly  from  the  meaning  of  the  word  equilibrium, 
which  implies  that  the  body  is  at  rest,  and  that  condition  can 
only  result  when  there  is  no  tendency  to  turn  the  body  about 
the  point  0;  that  is,  when  the  algebraic  sum  of  the  moments 
about  0  is  zero.  Should  the  line  of  action  of  a  force  pass 


CENTER  OF   GRAVITY  3 

through  the  center  of  moments,  the  moment  of  that  force  would 
vanish. 

In  expressing  the  value  of  moments,  the  units  of  force,  mass, 
area,  and  volume  are  placed  first  and  the  length  units  afterward. 
For  example,  a  moment  may  be  expressed  as  so  many  pounds- 
feet,  and  thus  avoid  confusion  with  work  units. 

3.  Center  of  Gravity.  —  The  center  of  gravity  of  a  body  or 
of  a  system  of  bodies  is  a  point  on  which  the  body  or  system 
will  balance  in  all  positions,  supposing  the  point  to  be  sup- 
ported, the  body  or  system  to  be  acted  on  only  by  gravity,  and 
the  parts  of  the  body  or  system  to  be  rigidly  connected  to  the 
point. 

It  follows  from  this  definition  that  all  the  particles  of  a  body 
or  of  a  system  of  bodies  are  acted  on  by  a  system  of  parallel 
forces,  gravity  acting  on  each,  and  that  the  algebraic  sum  of 
the  moments  of  these  forces  about  a  line  must  be  zero  when  the 
line  passes  through  the  center  of  gravity  of  the  body  or  system; 
otherwise  the  body  or  system  would  not  balance. 

Let  two  heavy  weights  P  and  Q  be  situated  as  shown  in  Fig.  2. 

Join  them  with  a  straight  line  and  divide  it  at  C  so  that  —  =  -  • 

If  the  weights  be  joined  by  a  rigid  rod  without  weight,  the 

system   will,  by  the  principle  of   the 

lever,  balance  when  supported  at  C. 

The  center  of  gravity  of  the  system  is 

therefore   at  C.     As  the  resultant  of 

the   weights   is    P  +  <2,    the   pressure 

on  the  support  will  be  P  +  Q. 

The  center  of  gravity  of  a  uniform  straight  rod  is  evidently 
at  the  middle  point  of  its  length,  and  when  considering  a  body 
at  rest  we  may  assume  its  whole  mass  to  be  concentrated  at 
its  center  of  gravity. 

Let  the  weights  P  and  Q,  Fig.  3,  be  attached  to  a  balanced 


4  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

rod   as   shown.     Then,  as  we  have  just  seen,   the  center  of 

P      n 

gravity  will  be  so  situated  that  —  =  — ,  or  Pm  =  Qn;  that  is, 

\l      m 
the  algebraic  sum  of  the  moments  is  zero. 

The  distance  of  the  center  of  gravity  of  a  system  from  a 
point  or  from  a  line  may  readily  be  determined  by  moments. 
Thus,  in  Fig.  3,  let  x  denote  the  distance  of  the  center  of  grav- 
ity of  the  system  from  0,  and  let  p  and  q  be  the  distances, 


Fig.  3- 


respectively,  of  the  centers  of  gravity  of  P  and  Q  from  0. 
We  have  from  the  figure,  m  =  x  —  p  and  n  =  q  —  x.  Since 
the  center  of  gravity  must  be  between  P  and  Q,  and  since  the 
turning  tendency  of  P  about  the  center  of  gravity  is  contra- 
clockwise  and  that  of  Q  clockwise,  we  shall  have  by  the  prin- 
ciple of  moments 

P(x-p)=Q(q-x), 

whence 

Pp+Qq 

P+Q 

If  the  system  be  extended  the  result 

_  _  Pp  +  Qq  +  Rr  +  etc. 
P  +  Q  +  R  +  etc.    '' 

will  be  obtained.  This  formula  is  extensively  used  in  the 
solution  of  problems. 

If  a  sheet  of  uniform  thickness  weighing  M  pounds  per  unit  of 
area  be  considered,  the  weight  of  any  area  a  will  be  Ma  pounds, 


CENTER  OF   GRAVITY 


and  we  may  substitute  Mai  for  P  and  Moz  for  Q  and  obtain 

_  _  Maip  +  Moyq  _  dip  +  #2<7  _  dip  +  a^q 
Mai  +  Mo2  #1  +  02  ^4 

in  which  ^4  is  the  whole  area.     This  may  be  expressed  as  follows: 
The  distance,  x,  from  O  to  the  center  of  gravity  is 

Sum  of  the  moments  of  all  the  elemental  surfaces  about  O 

/£  __ __ . 

Whole  surface 
_  Moment  of  the  whole  surface  about  O 

Whole  surface 
Similarly, 

Moment  of  the  whole  volume  about  0 

/y»     —     . ,     0 

Whole  volume 

The  moment  of  the  whole  surface  and  of  the  whole  volume  is 
obtained  by  integration. 

4.  The  center  of  gravity  of  a  portion  of  the  perimeter  of  a 
regular  polygon  or  of  an  arc  of  a  circle,  considered  as  a  thin  wire, 
may  be  obtained  thus: 

In  Fig.  4,  let  L  =  the  sum  of  the 
lengths  of  the  sides  of  the  portion  of  the 
perimeter  taken,  or  the  length  of  the  arc 
in  the  case  of  a  circle;  R  =  the  radius 
of  the  inscribed  circle,  or  the  radius  of 
the  circle;  C  =  the  chord  of  the  arc  of 
polygon  or  circle;  Y=  the  distance  of  the  center  of  gravity 
from  the  center  of  the  circle.  Then,  in  Fig.  5,  let  a  denote  the 


6  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

length  of  a  side  of  the  polygon.  The  center  of  gravity  of  each 
side  will  be  at  its  middle  point,  and  distant  yi,  yZ)  y%,  etc.,  from 
the  diameter  of  the  inscribed  circle.  Let  xi,  Xz,  #3,  etc.,  denote 
the  projected  lengths  of  the  sides  on  the  diameter.  From  the 
similar  triangles  Obc  and  edf  we  have 

de  :  Ob  =  df  :  be,     or    a  :  R  =  x\  :  yi, 

Rxi 

whence  y\  =  --- 

a 

o-    M    i  Roc?, 

Similarly,  y2  =  --  >  and  so  on. 

If  w  denotes  the  weight  of  each  of  the  n  sides  of  the  polygon, 
we  shall  have 


Y  =  Pp  +  Qq  +  etc.  =  wyi  +  wy2  +  etc.  =  yi  +  y2  +  etc.  ^ 
P  +  Q  +  etc.  nw  n 

Substituting  the  values  of  yi,  y^  ys,  etc.,  we  have 

Y  =  —  (*i  +  ai  +  etc.). 

na 

But         '  na  =  L,     and    x\  +  x^  +  etc.  =  C. 

Hence  Y  =  ^f- 

JL/ 

5.  The  center  of  gravity  of  a  plane  figure  may  be  obtained 
graphically  as  follows: 

Let  abcde,  Fig.  6,  be  any  plane  figure.  Draw  eb  and  ec.  Let 
Ai  and  gi,  At  and  #2,  and  As  and  g3  denote  the  areas  and  centers 
of  gravity  of  the  triangles  eab,  ebc,  and  ecd  respectively.  Join  gL 
and  gz.  The  center  of  gravity  of  the  figure  abce  must  lie  in  this 
line.  From  gi  lay  off  in  any  direction  and  to  any  scale  the  dis- 
tance git  equal  to  A^  and  in  the  opposite  direction  and  to  the 
same  scale  lay  off  from  g2  the  distance  g2S  equal  to  A  i  and  paral- 
lel to  git.  Join  st;  then  its  intersection,  k,  with  gig2  is  the  center 


CENTER   OF    GRAVITY 


of  gravity  of  the  figure  abce.  Join  k  and  g3.  The  center  of 
gravity  of  A\  +  A2  +  A3  will  lie  in  this  line.  From  k  lay  off 
in  any  direction  a  distance 
kr  equal  to  the  area  As,  and 
from  g3  lay  off  gsv  parallel 
to  kr  and  equal  to  A\  +  A2. 
Join  w,  and  its  intersection 
with  kgs  gives  G  as  the  cen- 
ter of  gravity  of  the  whole 
figure. 

Should  the  surface  con- 
tain a  hole,  Sisfmn  in  Fig.  7, 
we  would  proceed  thus: 

Denote  by  A\  and  gi 
and  A 2  and  £2  the  areas 
and  centers  of  gravity  of  Fig-  6- 

the  whole  figure  abcde  and  of  /ww  respectively.     Draw 

b 


From  £2  lay  off  to  scale,  in  any  direction,  g2s  equal  to  AI,  and 
parallel  to  it,  and  on  the  same  side  of  gigz,  lay  off  g\r.     Join 


8 


THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 


sr  and  extend  it  to  meet  gig2  produced  at  G.     Then  the  point  G 

is  the  required  center  of  gravity. 

The  truth  of  these  graphic  methods  may  be  shown  as  follows: 
Let  Oi  and  02,  Fig.  8,  be  the  positions  of  the  centers  of  gravity 

of   two    areas  A\  and   A2  respectively,  and   let  their  common 
d 


Fig.  8. 

center  of  gravity  be  situated  at  g,  distant  r\  from  Oi  and  r% 
from  02.  By  the  principle  of  moments  we  shall  have  A^r\  — 
A2r2.  In  any  direction  from  O2  lay  off  O2b,  whose  length  repre- 
sents the  area  A\  to  some  selected  scale,  and  from  O\  lay  off 
Old  parallel  to  02b  and  of  such  length  as  to  represent  the  area  A2 
to  the  same  scale.  The  triangles  Oidg  and  02bg  are  similar,  and 

we  have 

O2b  :  Old  =  O2g  :  Oig,     or     AI  :  A2  =  r2  :  ri} 

whence  A&I  =  A2r2. 

That  is,  the  line   joining  b  and  d  passes  through  g.     It  will  be 

observed  that  the  lines  O2b  and  Old  are  laid  off  on  opposite  sides 

of  the  line  joining  Oi  and  02  and 
at  opposite  ends  of  their  respec- 
tive areas  and  at  any  convenient 
angle.  Should  one  of  the  areas, 
as  A2,  be  negative,  i.e.,  the  area 
of  a  hole,  or  of  a  part  cut  out 

of  the  surface,  then  02b  and  Old  must  be  laid  off  on  the  same 

side  of  Oi02,  as  shown  in  Fig.  9. 


Fig.  9. 


CENTER   OF   GRAVITY  9 

6.  Applications.  —  To  show  the  application  of  the  foregoing 
principles  to  finding  centers  of  gravity,  the  solutions  of  a  few 
problems  will  be  given. 

EXAMPLE  I.  —  Find  the  center  of 
gravity  of  a  triangle. 

Conceive  the  triangle,  Fig.  10,  to 
be  divided  into  a  great  number  of 
very  narrow  strips  drawn  parallel  to 
one  of  the  sides,  as  B.  The  center 
of  gravity  of  each  strip  will  be  at  its 

middle  point,  therefore  the  center  of  gravity  of  the  triangle 
will  lie  in  the  locus  of  these  middle  points;  that  is,  in  the 
median  Oc. 

The  elemental  area  of  the  triangle  =  b  •  dh. 

The  moment  of  the  elemental  area  =  h*b  •  dh. 

From  similar  triangles  we  have 

b  :  B  =  h  :  H\  whence,  b  =  -  —  • 

ti 

Then 

B-h2-  dh 


Moment  of  elemental  area  = 


H 


7?    CH               7?  77^ 
Moment  of  the  whole  area  =  —   /      h2dh  = 


7?  TJ 

The  area  of   the  triangle  =  - 


The  distance,  x,  of  the  center  of  gravity  from  the  apex  is 

BH2 

-  _  Moment  of  whole  area  _     3      _  2  H 
Whole  area  ~  BH  '      3 


That  is,  the  center  of  gravity  is  at  a  perpendicular  distance 
below  the  apex  equal  to  two-thirds  of  the  altitude,  and  must 


10 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


therefore  be  at  g,  the  intersection  of  the  median,  Oc,  and  the 

parallel  to  the  base  distant  -  -  from  the  apex.     From  similar 

o 
triangles  it  is  seen  that  Og  is  two- thirds  the  length  of  the  median, 

so  that  the  center  of  gravity  is  on  the  median  at  two-thirds  its 
length  from  the  apex. 

EXAMPLE  II.  —  To  find  the  center  of  gravity  of  a  portion  of 
a  regular  polygon  or  of  a  sector  of  a  circle  when  considered  as 
a  lamina  or  thin  sheet. 


Fig.  ii. 

Referring  to  Fig.  n,  and  using  the  same  notation  as  in  Art.  4, 
we  shall  have  -   -  as  the  distance  of  the  center  of  gravity  of 

o 

each  of  the  triangles  from  the  center  of  the  inscribed  circle,  and 
they  will  be  distant  3/1,  y2)  y^,  etc.,  from  the  diameter.  The  base 
of  each  of  the  triangles  is  a,  and  the  projected  lengths  of  these 
bases  on  the  horizontal  diameter  are  Xi,  #2,  #3,  etc.  From  the 
similar  triangles  Obc  and  edf  we  have 


2  R 


:  a  = 


similarly, 


, 

,  whence  y 

, 
»  and  so  on. 


If  w  denotes  the  weight  of  each  of  the  n  triangles  of  the  polygon 
we  shall  have 

r_wyi  +  w3fr  +  etc._yi  +  3*  +  etc.=  2fl^  [  ^ ^  ctc^2  RC 
nw  n  $na  '          3  L 


CENTER  OF   GRAVITY 


II 


EXAMPLE  III.  —  A  square  is  divided  into  four  equal  triangles 
by  diagonals  intersecting  at  0;  if  one  triangle  be  removed,  find 
the  center  of  gravity  of  the  figure  formed 
by  the  three  remaining  triangles. 

Let  w  denote  the  weight  of  each  of  the 
remaining  triangles,  and  let  a  denote  the 
side  of  the  square,  Fig.  12.  The  weight 
2  w  of  the  side  triangles  may  be  supposed 
concentrated  at  their  common  center  of 
gravity  0.  The  distance  of  the  center  of 
gravity  of  the  lower  triangle  from  0  is 


Fig.  12. 


-  X  -  =  -  >  by  Example  I. 
233 


Then  if  x  denotes  the  distance  of  the 


center  of  gravity  of  the  three  remaining  triangles  from  0,  we  shall 
have 


_  =  Pp+Qg  = 
P  +Q 


a 
- 
3  =  at 

9 


2w  +  w 

That  is,  the  required  center  of  gravity  is  distant  one-ninth  the 

side  of  the  square  from  the  center  of  the  square. 

EXAMPLE  IV.  —  A  quarter  of  the  area  of  a  triangle  is  cut  off 

by  a  line  drawn  parallel  to  the  base.     Find  the  center  of  gravity 

of  the  remaining  quadrilateral. 

Let  EF  be  parallel  to  the  base  EC, 
Fig.  13,  and  let  the  triangle  AEF  be 
the  part  cut  off.  The  required  cen- 
ter of  gravity  will  lie  in  the  median 
AD.  The  triangles  AEF  and  ABC 
are  similar,  and  are  to  each  other 
in  area  as  i  14.  Since  the  areas  of 

similar  triangles  are  to  each  other  as  the  squares  of  homologous 

sides,  we  have 

—  • 

2 


Fig.  13. 


:  ~AD2  = 


whence    AG  = 


12 


THE   ELEMENTS  OF  MECHANICS   OF  MATERIALS 


Let  x  denote  the  distance  of  the  required  center  of  gravity  of 
the  quadrilateral  BEFC  from  A.     Then 


2XAD_         AD 

~"  2 


IAD. 


P-Q  4-i  9  9 

That  is,  the  required  center  of  gravity  is  in  the  median  AD  and 

at  seven-ninths  of  its  length  from  A . 

EXAMPLE  V.  —  Find  the  center  of  gravity  of  a  cone. 
Conceive  the  cone  of  Fig.  14  to  be  made  up  of  a  great  number 

of  thin  sections,  each  parallel  to  the  base.     The  center  of  gravity 

of  each  of  these  sections  will  be 
at  its  center,  therefore  the  center 
of  gravity  of  the  cone  will  lie  in 
the  locus  of  the  centers  of  these 
sections;  that  is,  it  will  lie  in 
the  line  joining  the  vertex  with 
the  center  of  gravity  of  the 
base. 

The   elemental   volume   of    the 
cone  is  irr^dh,  and  the  moment  of 

the  elemental  volume  is  irr*hdh.     From  similar  triangles  we  have 


Fig.  14. 


R:r=  H: 


Rh 
whence  r  =  -— 

£2 


Then  by  substitution  we  have 

Moment  of  elemental  volume  = 


Moment  of  whole  volume 


Volume  of  cone 


If  x  denotes  the  distance  of  the  center  of  gravity  of  the  cone 
from  the  vertex,  we  shall  have 


CENTER  OF   GRAVITY  13 


_  _  Moment  of  whole  volume  _       4      _  3  H 
Whole  volume  irR2H        A 

3 

That  is,  the  center  of  gravity  is  at  a  perpendicular  distance 
below  the  apex  equal  to  three-fourths  of  the  altitude,  and  must 
therefore  be  at  g,  the  intersection  of  the  line  Oc  joining  the  apex 
with  the  center  of  gravity  of  the  base  and  the  parallel  to  the  base 

distant  —  -  from  the  apex.     From  similar  triangles  it  is  seen 
4 

that  Og  —  -  —  ,  so  that  the  center  of  gravity  of  the  cone  is  in 

4 

the  line  joining  the  vertex  with  the  center  of  the  base  and  at 
three-fourths  its  length  from  the  vertex. 

EXAMPLE  VI.  —  Find  the  center  of  gravity  of  a  semicircular 
arc,  or  wire. 


Fig.  15. 

r>/~< 

In  Fig.  15  we  have,  from  Art.  4,  F  =  — ,  in  which  Y  denotes 

the  distance  of  the  center  of  gravity  from  the  center,  L  the 
length  of  the  arc,  and  C  the  chord  of  the  arc. 
Then 

F=  RX2R^2R^D 

TVR  7T  7T 


14  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

PROBLEMS 

1.  A  rod  3  feet  long  and  weighing  4  pounds  has  a  weight  of  2  pounds 
placed  at  one  end;  find  the  center  of  gravity  of  the  system. 

Ans.   One  foot  from  weighted  end. 

2.  Find  the  center  of  gravity  of  a  uniform  circular  disc  out  of  which 
another  circular  disc  has  been  cut,  the  latter  being  described  on  a  radius  of 
ihe  former  as  a  diameter. 

v^i  Ans.   One-sixth  of  radius  of  large  circle  from  center. 

3.  A  heavy  bar  14  feet  long  is  bent  into  a  right  angle  so  that  the  lengths 
of  the  portions  which  meet  at  the  angle  are  8  feet  and  6  feet  respectively; 
show  that  the  distance  of  the  center  of  gravity  of  the  bar  so  bent  from  the 
point  of  the  bar  which  was  the  center  of  gravity  when  the  bar  was  straight, 

is  ^  feet. 
7 

4.  The  middle  points  of  two  adjacent  sides  of  a  square  are  joined  and 
the  triangle  formed  by  this  straight  line  and  the  edges  is  cut  off;  find  the 
center  of  gravity  of  the  remainder  of  the  square. 

Ans.    z\  of  diagonal  of  square  from  center  of  square. 

5.  A  piece  of  uniform  wire  is  bent  into  the  shape  of  an  isosceles  tri- 
angle; each  of  the  equal  sides  is  5  feet  long,  and  the  other  side  is  8  feet  long; 
find  the  center  of  gravity.  Ans.   ^  of  the  altitude  from  the  base. 

6.  Find  the  center  of  gravity  of  a  figure  consisting  of  an  equilateral 
triangle  and  a  square,  the  base  of  the  triangle  coinciding  with  one  of  the 
sides  of  the  square. 

Ans.   At  a  distance  from  the  base  of  the  triangle  equal  to 


3    4~ 


26 
times  the  base. 

7.  A  table  whose  top  is  in  the  form  of  a  right-angled  isosceles  triangle, 
the  equal  sides  of  which  are  3  feet  in  length,  is  supported  by  three  vertical 
legs  placed  at  the  corners;  a  weight  of  20  pounds  is  placed  on  the  table  at  a 
point  distant  15  inches  from  each  of  the  equal  sides;  find  the  resultant 
pressure  on  each  leg.  Ans.   8£,  8f  ,  3!  Ibs. 

8.  A  BCD  is  a  quadrilateral  figure  such  that  the  sides  AB,  AD,  and  the 
diagonal  AC  are  equal,  and  also  sides  CB  a'nd  CD  are  equal;  find  its  center 
of  gravity. 

2  (ft  J-  }? 

Ans.  -  ^-—  units  from  C,  in  which  a  =  AB  and  b  =  CB. 
6a 

9.  ABC  represents  a  triangular  board  weighing  10  pounds.     Suppose 
weights  of  5  pounds,  5  pounds,  and  10  pounds  are  placed  at  A,  B,  and  C 
respectively.    Where  is  the  center  of  gravity  of  the  whole  ? 

Ans.  At  five-ninths  of  the  median  drawn  from  C. 


CENTER   OF   GRAVITY  15 

10.  A  rod  of  uniform  thickness  is  made  up  of  equal  lengths  of  three 
substances,  the  densities  of  which  taken  in  order  are  in  the  proportion  of 
i,  2,  and  3;  find  the  position  of  the  center  of  gravity  of  the  rod. 

Ans.   At  seven-eighteenths  of  the  length  of  the  rod  from  the  end  of  the 
densest  part. 

11.  Find  the  position  of  the  center  of  gravity  of  a  piece  of  wire  bent  to 
form  three-fourths  of  the  circumference  of  the  circle  of  radius  R. 

Ans.   On  a  line  drawn  from  the  center  of  the  circle  to  a  point  bisecting 
the  arc,  and  at  a  distance  0.3  R  from  the  center. 

12.  A  thin  wire  forms  an  arc  of  a  circle  the  radius  of  which  is  10  inches, 
and  subtends  an  angle  of  60°;  find  the  distance  of  the  center  of  gravity 

from  the  center.  Ans.   —  ins. 


7T 


13.  Find  the  position  of  the  center  of  gravity  of  a  balanced  weight 
having  the  form  of  a  circular  sector  of  radius  R,  subtending  an  angle  of 
90°.  Ans.   0.6  R  from  center  of  circle. 

14.  Find  the  center  of  gravity  of  a  semicircular  lamina,  or  sheet,  of 

diameter  D.  Ans.   -  -  from  the  center. 

3^r 

15.  A  square  stands  on  a  horizontal  plane;  if  equal  portions  be  removed 
from  two  opposite  corners  by  straight  lines  parallel  to  a  diagonal,  find  the 
least  portion  which  can  be  left  so  as  not  to  topple  over. 

Ans.   Three-quarters  of  the  area  of  the  square. 
1  6.   Find  the  center  of  gravity  of  a  trapezoid. 
Ans.   On  the  line  joining  the  middle  points  of  the  bases,  and  at  a  per- 

pendicular distance  from  the  upper  base  equal  to  —  •  —  —  —  ,  and  from  the 

3      B  +  b 

lower  base,  —  -  2^  "*"     ,  B  and  b  being  the  lower  and  upper  bases  respec- 
3      B  +  o 


tively,  and  H  the  altitude.     If  the  distance  be  measured  on  the  line  S  joining 

01  TT 

the  middle  points  of  the  bases,  then  -  must  be  substituted  for  —  - 

3  «j 

17.   Find  the  center  of  gravity  of  a  pyramid. 

Ans.   In  the  line  joining  the  vertex  with  the  center  of  gravity  of  the 
base  and  at  three-fourths  its  length  from  the  vertex. 

1  8.   Find  the  center  of  gravity  of  a  frustum  of  a  cone. 

Ans.   In  the  line  joining  the  centers  of  gravity  of  the  upper  and  lower 

bases,  at  a  distance  from  the  upper  base  equal  to  —  ^—=^  -  =—  —  ^—  ; 

4         R   ~T~  Rf  ~T"  T 

and  from  the  lower  base  —  •  3  L>+  2^  +  »    »  R  and  r  bein§  the  radii  °* 
4       R  +  Rr  +  r 


l6  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

the  lower  and  upper  bases  respectively,  and  H  the  altitude.  These 
results  are  true  for  the  frustum  of  any  pyramid  by  substituting  B  for  R 
and  b  for  r,  B  and  b  being  homologous  sides  of  the  lower  and  upper  bases 
respectively. 

19.  A  lever  safety  valve  is  required  to  blow  off  at  70  pounds  pressure 
per  square  inch.     Diameter  of  valve,  3  inches;  weight  of  valve,  3  pounds; 
short  arm  of  lever,   2.5  inches;  weight  of  lever,   n  pounds;  distance  of 
center  of  gravity  of  lever  from  fulcrum,  15  inches.     Find  the  distance  at 
which  a  cast-iron  ball  6  inches  in  diameter  must  be  placed  from  the  ful- 
crum.   The  weight  of  the  ball-hook  is  0.6  pound,  and  that  of  a  cubic  inch 
of  cast  iron  0.26  Ib.  Ans.   35.48  ins. 

20.  A  steel  safety-valve  lever  i  inch  thick  and  50  inches  long  tapers 
from  3  inches  in  depth  at  the  fulcrum  to  i  inch  at  the  end.     It  overhangs 
the  fulcrum  4  inches,  the  overhang  having  no  taper.     Diameter  of  valve, 
4  inches;   weight  of  valve,  4.75  pounds;   short  arm  of  lever,  3.5  inches. 
Find  the  distance  from  the  fulcrum  at  which  a  cast-iron  ball  9.5  inches  in 
diameter  must  be  placed  in  order  that  steam  shall  blow  off  at  125  pounds 
pressure  per  square  inch.     Weight  of  the  ball-hook,  1.3  pounds;  weight  of 
a  cubic  inch  of  steel,  0.28  pound;  weight  of  a  cubic  inch  of  cast  iron,  0.26 
pound.  Ans.   42.35  ins. 

21.  A  trapezoidal  wall  has  a  vertical  back  and  a  sloping  front  face; 
width  of  base,  10  feet;  width  of  top,  7  feet;  height,  30  feet.     What  hori- 
zontal force  must  be  applied  at  a  point  20  feet  from  the  top  in  order  to 
overturn  it,  i.e.}  to  make  it  pivot  about  the  toe?     Width  of  wall,  i  foot; 
weight  of  masonry  in  wall,  130  pounds  per  cubic  foot. 

Ans.   18,900  Ibs. 

22.  Find  the  height  of  the  center  of  gravity  from  the  base  of  a  column 
4  feet  square  and  40  feet  high,  resting  on  a  tapered  base  forming  a  frustum 
of  a  square-based  pyramid  10  feet  high  and  8  feet  square  at  the  base. 

Ans.   20.4  feet  from  base 

Problems  23,  24,  and  25  are  to  be  solved  graphically. 

23.  Find  the  position  of  the  center  of  gravity  of  an  unequally  flanged 
section;  top  flange,  3  inches  wide,  1.5  inches  thick;  bottom  flange,  15 
inches  wide,  1.75  inches  thick;  web,  1.5  inches  thick;  total  height,  18  inches. 

Ans.    5.72  inches  from  the  bottom  edge. 

24.  Find  the  height  of  the  center  of  gravity  of  a  T-section  from  the 
foot,  the  top  crosspiece  being  12  inches  wide  and  4  inches  deep;  the  stem, 
3  feet  deep  and  3  inches  wide. 

Ans.   24.16  inches.     (Check  the  result  by  seeing  if  the  moments  are 
equal  about  a  line  passing  through  the  section  at  the  height  found.) 


MOMENT   OF   INERTIA  17 

25.  A  square  board  weighs  4  pounds,  and  a  weight  of  2  pounds  is  placed 
at  one  of  the  corners.  Find  the  position  of  the  center  of  gravity  of  the 
board  and  weight. 

Ans.  On  the  diagonal  drawn  from  the  weighted  corner  and  at  two- 
thirds  its  length  from  the  opposite  corner. 

7.  Moment  of  Inertia.  —  The  moment  of  inertia  of  a  surface 
is  the  sum  of  the  products  of  each  elemental  area  of  the  sur- 
face by  the  square  of  its  distance  from  an  axis  about  which  the 
surface  is  supposed  to  be  revolving.  If,  instead  of  a  surface, 
we  have  a  body,  then  we  must  substitute  the  elemental  volume 
for  the  elemental  area  in  finding  the  moment  of  inertia.  The 
moment  of  inertia  varies  according  to  the  position  of  the  axis, 
being  smallest  when  the  axis  passes  through  the  center  of 
gravity. 

It  would  be  a  crude  and  inaccurate  method  of  finding  the 
moment  of  inertia  by  actually  dividing  an  area  or  a  volume 
into  its  elements,  multiplying  each  by  the  square  of  its  distance 
from  the  axis,  and  then  finally  taking  the  sum  of  these  products. 
We  can,  however,  find  the  moment  of  inertia  with  accuracy  by 
means  of  integration. 

The  strength  of  a  beam  or  of  a  column  depends  upon  the  form 
as  well  as  the  area  of  its  section,  and  in  all  calculations  respecting 
the  strength  of  beams  and  columns,  the  factor  which  gives 
expression  to  the  effect  of  the  form  of  the  section  is  its  moment 
of  inertia  about  an  axis  passing  through  the  center  of  gravity. 

The  moment  of  inertia  of  a  surface  is  universally  denoted  by 
/,  the  axis  being  in  the  plane  of  the  surface  and  passing  through 
its  center  of  gravity.  When  the  axis  is  perpendicular  to  the 
plane  of  the  surface,  the  moment  of  inertia  is  then  termed  the 
polar  moment  of  inertia,  and  is  denoted  by  Ip. 

If  we  denote  the  product  of  a  force,  area,  volume,  or  weight  by 
its  arm  as  the  first  moment,  or  simply  the  moment,  of  the  force, 
area,  volume,  or  weight,  then  we  may  conveniently  denote  the 


i8 


THE    ELEMENTS   OF  MECHANICS  OF   MATERIALS 


product  of  the  force,  area,  volume,  or  weight  by  the  square  of 

its  arm  as  the  second  moment  of  the  force,  area,  volume,  or  weight. 

Thus  the  moment  of  inertia  is  sometimes  known  as  the  second 

moment. 

8.   Relation   between   Moments   of    Inertia   about    Parallel 

Axes.  —  There  is  a  relation  between  moments  of  inertia  of  a 

surface  about  parallel  axes  that  is  useful  in  the  solution  of 

problems. 

Let  I  denote  the  moment  of  inertia  of  any  surface  about  an 

axis  through  the  center  of  gravity 
of  the  surface,  /'  the  moment  of 
inertia  of  the  same  surface  about 
any  other  parallel  axis,  A  the  area 
of  the  surface,  and  D  the  per- 
pendicular distance  between  the 
axes.  Then  we  shall  have  /'  =  /  + 


For,  let  yy,  Fig.  16,  be  the  axis 
through  the  center  of  gravity,  and 
y'y'  the   axis  parallel  to  yy,   both 
being  in  the  plane  of  the  surface. 
Denote  the  elemental  areas  of  the  surface  by  0i,  02,  03,  etc., 
and  their  distances  from  the  axis  yy  by  r\,  r2,  rs,  etc. 
Then 

I'  =  ai(ri  -  D)2  +  02  (D  -  r2)2  +  az  (D  +  rs)2  +  etc. 


r32) 


-  2 
etc. 


—  2  D 


+ 


The  first  term  of  the  second  member  of  this  equation  is,  by 
our  definition,  the  moment  of  inertia,  7,  of  the  surface  about 
the  axis  yy  through  the  center  of  gravity;  the  second  term 


RADIUS    OF    GYRATION 


Fig.  17- 


becomes  AD2,  in  which  A  denotes  the  whole  area;  and  the  third 
term  will  reduce  to  zero,  because  the  quantity  within  the  paren- 
thesis is  the  algebraic  sum  of  the  moments  of  the  elemental 
areas  about  a  line  passing  through  their  center  of  gravity. 
Therefore  I'  =  I  +  AD2. 

9.  Suppose  an  axis  perpendicular  to  the  plane  of  the  surface 
represented  in  Fig.  17  to  pass  through  0.     Let  a  be  an  elemental 
area  of  the  surface,  distant  r  from  0, 

and  let  X  and  Y  be  rectangular  axes 

passing  through  O  and  lying  in  the 

plane  of  the  surface.    Then  the  polar 

moment  of  a  is  Ip  =  ar2.  The  moment 

of  inertia  of   a  about  X  is  Ix  =  ay2, 

and  similarly  /„  =  ax2.    But  we  have 

x2  +  y2  =  r2,  therefore  ax2  +  ay2  =  ar2. 

That  is,  the  polar  moment  of  inertia 

of  any  surface  is  equal  to  the  sum  of 

the  moments  of  inertia  of  the  surface  about  any  two  rectangular 

axes  lying  in  the  plane  of  the  surface  and  passing  through  the 

polar  axis  of  revolution ;  that  is, 

/,-/,  +  /„. 

10.  Radius  of  Gyration.  —  We  have  seen,  Art.  8,  that  /  = 
a\r\  +  <W2  +  a3r32  +  etc.,  in  which  #1  -f-  #2  +  #3  +  etc.  =  ^> 
the  whole  area.     If  we  can  conceive  the  whole  area  to  be  con- 
densed into  a  single  particle,  distant  K  from  the  axis  of  rotation, 
we  shall  have  /  =  AK2.     This  imaginary  point  at  which  the 
particle  is  supposed  to  be  situated  is  known  as  the  center  of 
gyration,  and  its  distance,  K,  from  the  axis  is  the  radius  of 

gyration.     We  have  then,  K  =  y  —  >  in  which  the  mass,  M,  the 

volume,  V,  or  the  weight,  W,  may  be  substituted  for  the  area,  A. 
u.  Illustrations.  —  To  illustrate  the  methods  of  finding  mo- 
ments of  inertia  and  radii  of  gyration  a  few  examples  follow. 


2O 


THE    ELEMENTS   OF  MECHANICS  OF   MATERIALS 


EXAMPLE  I.  —  To  find  the  least  moment  of  inertia  and  least 
radius  of  gyration  of  the  surface  of  a  parallelogram. 

The  least  moment  of  inertia  is  that  about  an  axis  passing 
through  the  center  of  gravity  and  parallel  to  the  bases.  Let  X  X, 
Fig.  1 8,  be  the  axis  through  the  center  of  gravity  and  parallel 
to  the  base  B.  It  bisects  the  altitude  h. 


Fig.  18. 

Suppose  the  surface  to  be  divided  into  an  infinite  number  of 
strips  parallel  to  the  axis,  each  of  thickness  dy.  Consider  one 
of  these  strips  distant  y  from  the  axis. 

Then  Elemental  area  =  Bdy, 


Second  moment  of  elemental  area  =  dl  =  By2dy. 


Then 


24 


12 


If  the  axis  YY  parallel  to  the  base  b  be  taken,  we  shall  have 

bH* 


12 


RADIUS   OF   GYRATION 
For  the  radius  of  gyration  we  shall  have 


21 


Bh3        h 


and 


If  the  moment  of  inertia  with  respect  to  an  axis  coinciding 
with  the  base  be  desired,  we  have  I'  —  I  +  AD2,  in  which  D 
denotes  the  perpendicular  distance  between  the  axes,  denoted  in 

h  77 

this  instance  by-  when  B  is  the  axis,  and  by—  when  b  is  the  axis. 


I'  =  —  +  Bh  X  -  =  —for  the  base  B  as  axis, 
12  43 


and 


Also 


/'=:  —  + 

12 


X  --  =  - — •  for  the  base  b  as  axis. 
4          3 

£'  =  V  —77  =  -  V^  for  the  base  B  as  axis, 
v  3  7^      3 

O  vx 

_,      4  /l^"      77   /-  , 

C  =  y  —7—  =  —  v  3  for  the  base  6  as  axis. 


EXAMPLE  II.  —  Find  the  moment  of  inertia  and  radius  of  gyra- 
tion of  a  circular  surface   about  an   axis 
passing  through  the  center   and  perpen- 
dicular to  the  plane  of  the  surface. 

The  axis  being  perpendicular  to  the  sur- 
face it  is  the  polar  moment  of  inertia  that 
is  required. 

Conceive  the  circular  surface  of  Fig.  19 
to  be  made  up  of  an  infinite  number  of 
concentric  circular  strips,  each  of  thick- 
ness dr.  Consider  the  strip  distant  r  from  the  center. 

Then  Elemental  area  =  2  irrdr, 

Second  moment  of  elemental  area  =  dl  =  2 


22 


THE   ELEMENTS  OF  MECHANICS  OF   MATERIALS 


Then  I  = 


— 

2 


32 


J  *D*    = 

V  2- 


K= 

'         A 


EXAMPLE  III.  —  Find  the  polar  moment  of  inertia  and  radius 
of  gyration  of  a  right  circular  cone  about  its  axis. 

Conceive  the  volume  of  the  cone, 
Fig.  20,  to  be  made  up  of  an  infinite 
number  of  circular  strips,  each  of 
thickness  dh.  Consider  the  strip  of 
radius  r  and  distant  h  from  the 
vertex. 
Then 

Elemental  volume  =  irr2dh. 

Fls-  20-  If  we  imagine  the  elemental  volume 

to  be  condensed  into  a  single  particle,  the  distance  of   this 
particle  from  the  axis  will  be  the  radius  of  gyration  K,  which 

for  the  circle  is  -  —  -• 


Hence, 
Second  moment  of  elemental  volume 


dl,= 


Then 


*-r 

2*/n 


Hdr 


By  similar  triangles,  —  =-  >  whence  h  =  — ,  and  dh  =  —  • 
Then 

'.-fir*- 


RADIUS   OF    GYRATION  23 

PROBLEMS 

Find  the  moment  of  inertia  and  radius  of  gyration  of  the  following: 

1.  Triangle  about  an  axis  through  vertex  and  parallel  to  base. 

Ans.  7'-**;   K  =  *^. 

4     '  2 

2.  Triangle  about  an  axis  through  the  center  of  gravity  and  parallel 


to  base.  Ans.  I  =  -  -  ;    K  =  —  Vi8. 

36  18 

3,   Triangle  about  an  axis  coinciding  with  base. 

R  773  JJ       — 

Ans.  I 


12  6 

4.   Trapezoid  about  an  axis  coinciding  with  small  base. 


5.   Trapezoid  about  an  axis  coinciding  with  large  base. 

j,_fP(3b  +  B)_  /3b+B 

~~  B- 


6.   Trapezoid  about  an  axis  through  center  of  gravity  and  parallel  to 
base. 


7.  Square  about  its  diagonal. 

Ans.  I  =  a-  ;  K  =  ±  VT,. 

12  12 

8.  Circle  about  a  diameter. 

A  T        irD*       j-r        D 

Ans.  I  =  —  —  ;  K  =  —  • 
64  4 

9.  Hollow  circle  about  a  diameter. 

Ans.  /-f  (/>•-*);  *  =  VZ)2+tP. 
64  4 

Find  the  polar  moment  of  inertia  and  radius  of  gyration  of  the  fol 
lowing  named  surfaces: 

10.  Parallelogram  about  a  pole  passing  through  its  center  of  gravity. 


ii.  Hollow  circle  about  a  pole  passing  through  center. 

Ans.  /,.JL(o.-#); 

32 


24  THE   ELEMENTS   OF  MECHANICS  OF  MATERIALS 

Find  the  polar  moment  of  inertia  and  radius  of  gyration  of  the  follow- 
ing named  solids: 

12.  Cylinder  about  a  pole  coinciding  with  its  axis. 

Ans.  Ip  = 

32 

13.  Hollow  cylinder  about  a  pole  coinciding  with  its  axis. 

14.  Sphere  about  a  diameter 


5 

15.   A  bar  of  rectangular  section  about  a  pole  passing  through  the 
center  of  figure. 


CHAPTER  II 

BENDING    MOMENT.    BENDING    MOMENT    DIAGRAM. 
SHEAR.     SHEAR  DIAGRAM. 

12.  Bending  Moments.  —  When  forces  act  on  a  body  in  such 
a  manner  as  to  tend  to  give  it  a  spin  or  a  rotation  about  an  axis 
without  any  tendency  to  shift  its  center  of  gravity,  the  body  is 
said  to  be  acted  on  by  a  couple.  A  couple  consists  of  two  parallel 
forces  of  equal  magnitude  acting  in  opposite  directions  but  not 
in  the  same  line,  the  arm  of  the  couple  being  the  perpendicular 
distance  between  the  lines  of  action  of  the  forces. 

A  beam  is  subjected  to  a  bending  moment  when  it  is  so  acted 
upon  at  its  ends  by  equal  and  opposite  couples  that  there  is  a 
tendency  to  turn  it  in  opposite  directions. 


R  +  W 


Fig.  21. 

Thus,  the  beam  mn  of  Fig.  21  is  acted  on  by  the  equal  and 
opposite  couples,  R  and  R,  and  W  and  W,  the  tendency  being 
to  turn  the  beam  in  opposite  directions  about  the  point  r\  that 
is,  to  bend  it  at  r.  In  Fig.  22,  the  couples  whose  moments  are 
R\  X  mr  and  RI  X  nr  have  the  same  effect  on  the  beam  as  those 
of  Fig.  21.  The  beam  of  Fig.  21  is  called  a  cantilever,  from  the 

25 


26  THE   ELEMENTS    OF   MECHANICS  OF   MATERIALS 

nature  of  its  support,  while  the  beam  of  Fig.  2  2  is  called  a  simple 
beam.     The  forces  R,  RI  and  R2  are  the  support  reactions. 

W|=R1+R2 

CD 


Fig.  22. 

13.  General  Case  of  Bending  Moments.  —  The  bending 
moment  at  any  section  of  a  beam  is  the  algebraic  sum  of  the  moments 
of  all  the  external  forces  acting  to  the  left  of  the  section.  It  is 
assumed  that  forces  acting  upward  are  positive,  and  those  act- 
ing downward  are  negative,  so  that  bending  moments  may  be 
positive  or  negative. 

In  graphical  constructions  the  signs  of  bending  moments  are 
of  the  first  importance  and  are  determined  by  the  following 
rule: 

Bending  moments  which  tend  to  bend  a  beam  or  cantilever 
concave  upward,  ^ — -^  are  regarded  as  positive,  and  when  they 
tend  to  bend  in  the  reverse  way,  x- — ^,  they  are  negative. 

It  should  be  observed  that  it  is  merely  to  avoid  confusion  in 
the  construction  of  diagrams  that  the  external  forces  to  the  left 
of  the  section  were  considered  when  defining  the  bending  moment 
at  any  section  of  a  beam.  Moments  may  equally  well  be  taken 
to  the  right  of  the  section  and  the  same  value  be  obtained  for 
the  bending  moment  at  the  section.  When  bending  moments 
are  obtained  by  calculation  rather  than  by  construction,  ^the 
side  involving  the  least  calculation  in  taking  moments  should 
be  selected,  though  the  calculations  for  both  sides  afford  a  posi- 
tive check  as  to  the  accuracy  of  the  work.  To  avoid  confusion 
with  work  units,  bending  moments  are  expressed  in  pounds- 
inches,  pounds-feet,  or  tons-inches,  as  may  be  found  most  con- 
venient. 


BENDING  MOMENTS  27 

The  beam  of  Fig.  23  is  supposed  to  be  without  weight.  The 
bending  moment  at  section  E  is,  taking  moments  to  the  left 

of  E, 

M  =  Ri  X  aE  -  Wl  X  qE  -  W2  X  pE,  (i) 

or,  taking  moments  to  the  right  of  E, 

M  =  R2  X  bE  -  W3  X  rE.  (2) 

Suppose  the  distances  and  weights  to  be  as  shown  in  the 
figure.  Before  finding  the  bending  moments  we  must  first  find 
the -reactions,  RI  and  R2. 


Wf=  60  Ibs. 


>=50  Ibs. 


W,=  S01bs. 


<        »'         ilc                3'             sic     15'  s 

2' 

9' 

cb      el) 

.    C 

i 

q                            p                                 T                 * 

R!                                                                                                     RS 

Fig.  23. 

Taking  moments  about  the  left  support,  we  have 

#2  X  10.5  =  80  X  8.5  +  50  X  5  +  60  X  2, 
whence  RI  =  100  pounds. 

Taking  moments  about  the  right  support,  we  have 

R!  X  10.5  =  60  X  8.5  +  50  X  5-5  +  80  X  2, 
whence  RI  =  90  pounds, 

which  might  have  been  expected,  since  ^i  +  R2  should  equal 
Wi  +  W2  +  Wz. 

By  substitution  in   equation  (i),  we   have    for  the  bending 
moment  at  E, 

M  =  90  X  6.5  —  60  X  4.5  —  50  X  1.5  =  240  pounds-feet. 
Substituting  in  equation  (2), 

M  =  100  X  4  —  80  X  2  =  240  pounds-feet. 

It  is  thus  seen  that  the  bending  moment  at  a  section  is  the  same, 
whether  the  moments  be  taken  to  the  left  or  to  the  right  of  the 


28 


THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 


section.     In  this  instance  the  taking  of  moments  to  the  right 
involved  the  least  calculation. 

14.  Bending  Moment  Diagrams.  —  A  diagram  may  be  made 
to  show  graphically  the  bending  moment  at  any  section  of  a 
beam.  Such  diagrams  are  known  as  bending-moment  diagrams. 


Wx=601bs. 


W2=501bs. 


W  =801138. 


.     - 

c 

5  f 

}       C 

•^—  l.b—  ^  2  «  > 

)  .  C 

<:  z~  —  > 

„ 

RI 

o: 

p 

... 

r              t 

Fig.  24. 

For  example,  take  the  beam  of  Fig.  23.     The  bending  moment 

at  £  is 

M  =  RI  X  2  =  90  X  2  =  180  lbs.-ft; 

at  #  M  =  RI  X  5  -  TFi  X  3  =  90  X  5  -  60  X  3  =  270 lbs.-ft.; 
at  r   M  =  RI  X  8.5  -  Wi  X  6.5  -  TF2  X  3.5 

=  90  X  8.5  -  60  X  6.5  -  50  X  3.5  =  200  lbs.-ft. 

If,  on  a  base  line  a'b'y  Fig.  24,  and  to  a  scale  of  i  inch  =  200 
pounds-feet,  we  erect  ordinates  to  represent  these  bending 
moments,  and  then  join  their  extremities  with  the  broken  line 
a'q'p'r'b',  the  inclosed  figure  is  a  diagram  whose  ordinates 


SHEAR   DIAGRAMS  2Q 

beneath  any  section  of  the  beam  will  measure  the  bending 
moment  at  the  section  to  the  scale  adopted.  Thus  the  ordinate 
for  the  bending  moment  at  q  measures  iS$  =  0.9  inch;  that  at  p, 
J$S  =  1.35  inches;  and  that  at  r,  f§#  =  i  inch.  The  ordinate  be- 
neath E  measures  1.2  inches,  which,  to  scale,  represents  a  bend- 
ing moment  of  1.2  X  200  =  240  pounds-feet,  as  already  found  by 
calculation. 

15.  Shear.  —  Shearing   stresses   exist   when    couples,    acting 
like  a  pair  of  shears,  tend  to  cut  a  body  between  them.     Beams 
acted  on  by  couples  are  subjected  to  shearing  stresses  as  well  as 
to  bending  moments,  the  latter  being  far  more  important  in 
beams  of  lengths  ordinarily  encountered.     The  failure  of  very 
short  beams  will  be  invariably  from  shear. 

The  shear  at  any  section  of  a  beam  or  of  a  cantilever  is  equal  to 
the  algebraic  sum  of  the  forces  to  the  left  of  the  section,  the  upward 
forces  being  regarded  as  positive  and  the  downward  negative. 

16.  Shear  Diagrams.  —  Diagrams  made  to  show  graphically 
the  amount  of  the  shear  at  any  section  of  a  beam  are  known  as 
shear  diagrams.     In  their  construction,  attention  must  be  paid 
to  signs,   so  that  in  cases  where  the  shear  is  partly  positive 
and  partly  negative  the  positive  part  may  be  placed  above  the 
shear  axis,  or  base  line,  and  the  nega- 
tive part  below. 

The  failure  from  overload  of  a  short 
cantilever,  Fig.  25,  will  be  from  shear, 
the  projecting  part  shearing  off  bodily 
from  the  part  built  in.  According  to 
our  definition  the  shear  at  all  sections  of 
the  beam  is  constant  and  equal  to  —  W. 
The  shear  is  therefore  represented  Flg-  25< 

graphically  by  the  small  rectangular  shaded  diagram  of  Fig.  25, 
each  of  the  ordinates  of  which  is  equal  to  —W.  As  the  shear 
diagram  is  negative,  it  is  placed  below  the  shear  axis  mn. 


3° 


THE    ELEMENTS   OF  MECHANICS   OF  MATERIALS 


In  the  case  of  the  simple  beam  of  Fig.  26,  a  failure  from  shear 
will  occasion  the  part  between  the  supports  to  slide  down,  as 
shown  by  the  dotted  lines.  By  our  definition  the  shear  at 
any  section  to  the  left  of  the  load  W  is  RI,  and  at  any  section 
between  W  and  the  right  support  it  is  RI  —  W  =  —  R^  since 
R1  +  R2=W. 

The  shaded  part  of  Fig.  26  represents  the  shear,  and  shows 
that  it  changed  sign  under  the  load,  the  positive  part  being  placed 
above  the  shear  axis  mn  and  the  negative  part  below. 

As  in  the  case  when  "defining  bending  moments,  only  the 
external  forces  to  the  left  of  the  section  were  considered  when 
defining  the  shear  at  the  section,  but  this,  as  in  the  case  of 
bending  moments,  was  only  for  convenience.  If  Wa  denotes 
the  sum  of  all  the  loads  between  the  left  support  and  a  given 
section,  and  Waf  the  sum  of  all  the  loads  between  the  section 
and  the  right  support,  then  evidently  RI  -\-  R2  =  Ws  -}-  Wa', 
whence  Rl-W8=-  (R2  -  Wsr).  The  first  member  of  this 
equation  is  the  algebraic  sum  of  the  forces  to  the  left  of  the 

given  section  and  is  the  shear  at 
the  section,  and  the  second  member 
is  the  negative  of  the  algebraic  sum 
of  the  forces  to  the  right  of  the  sec- 
tion. The  shear  at  a  given  section 
is,  then,  the  algebraic  sum  of  the  ex- 
ternal forces  to  the  right  or  to  the 
left  of  the  section,  but  with  contrary 
signs. 

Let  the  beam  of  Fig.  27  be  loaded 
with  Wi,  Wij  W$  at  distances  di,  d2,  ds 

Fig.  26.  A.      ,      r  . 

respectively  from  any  given  section 

a.     Denote  the  support  reactions  by  RI  and  R2,  and  their  dis- 
tances from  a  by  r\  and  r2  respectively. 

To  construct  the  shear  diagram  of  the  beam  we  select  a  base 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  31 

line,  or  axis,  mn,  and  having  determined  to  an  appropriate  scale 
the  values  of  R\  and  R2  from  the  given  weights  TFi,  W2,  and  Wz, 
we  proceed  as  follows: 

By  our  definition  the  shear  at  any  section  to  the  left  of  TFi 
is  RI,  which  gives  the  point  b  of  Fig.   28.     Immediately  the 


w 


Fig.  29. 


point  of  application  of  W\  is  passed  the  shear  becomes  R\  —  Wi, 
giving  the  point  c,  and  this  shear  continues  unchanged  up  to 
the  point  of  application  of  Wi,  but  immediately  this  point  is 
passed  the  shear  becomes  R\  —  W\  —  W%,  giving  the  point  d, 


32  THE   ELEMENTS  OF  MECHANICS  OF   MATERIALS 

the  shear  becoming  negative.  This  shear  continues  unchanged 
until  the  point  of  application  of  W$  is  passed,  when  the  shear 
becomes  R\  —  W\  —  W%  —  Ws  =  —  RZ,  giving  the  point  e.  This 
shear  continues  unchanged  to  the  right  support,  giving  the  point 
o  and  completing  the  diagram. 

Figures  27  and  28  were  constructed  to  the  following  scales: 
Linear,  i  inch  =  4  feet;  load,  i  inch  =  150  pounds. 

The  data  of  the  beam  and  loads  were:  Length  of  beam  = 
12  feet  =  3  inches  to  scale;  di  =  2  feet  =  0.5  inch  to  scale; 
^2  =  3  feet  =  0.75  inch  to  scale;  d3  =  6  feet  =  1.5  inches  to 
scale;  r\  =  5  feet  =  1.25  inches  to  scale;  r2  =  7  feet.  =  1.75 
inches  to  scale;  W\  =  60  pounds  =  0.4  inch  to  scale;  W2  =  45 
pounds  =  0.3  inch  to  scale;  Ws  =  go  pounds  =  0.6  inch  to 
scale. 

The  support  reactions  were  first  found  as  follows: 

Taking  moments  about  the  left  support,  we  have 

R*  (ri  +  r2)  =  W3  fa  +  J8)  +  W2  fa  +  <fe)  +  Wl  fa  -  dj, 
or  12  R2  =  go  X  n  +  45  X  8  +  60  X  3, 

whence  R%  =  127.5  pounds. 

Therefore  RI  =  60  +  45  +  90  —  127.5  =  67.5  pounds. 

Reduced  to  scale  the  reactions  RI  and  R2  measure 

—  —  =  0.45  inch  and  —  "•  =  0.85  inch  respectively. 
15° 


The  bending-moment  diagram,  Fig.  29,  was  constructed  on  the 
base  line  m'n'  to  a  scale  such  that  i  inch  in  depth  represents 
300  pounds-feet.  The  bending  moments  at  Wi,  W2,  and  W$ 
were  calculated  as  follows: 

MWl=  Ri  fa  -  di)  =  67.5  X  3  =  202.5  pounds-feet  =  ^^ 

o 
=  0.675  mcn  t°  scale. 

Mw,=  Ri  fa  +  4)  -  Wi  (di  +  4)  =  67.5  X  8  -  60  X  5 


24.0  pounds-feet.  =  -—  =  0.8  inch  to  scale. 
300 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS 


33 


=  Ri_(ri  +  ds)  -  Wi  (fa  +  d3)  -  W2  (d3  -  <fe) 
=  67.5  X  ii  —  60  X8  —  45X3  =  127.5  pounds-feet. 
127-5 


300 


=  0.425  inch  to  scale. 


Ordinates  equal  in  length  to  the  scale  measurements  repre- 
senting these  bending  moments  were  erected  on  m'n' ,  as  shown, 
and  their  extremities  joined  by  the  broken  line  m'fghkn'.  The 
ordinate  of  the  resulting  diagram  beneath  any  section  of  the  beam 
is  the  scale  measurement  of  the  bending  moment  at  the  section. 
Thus  the  ordinate  under  the  section  at  a  measures  f  f  of  an  inch, 
corresponding  to  a  bending  moment  of  ff  X  300  =  217.5  pounds- 
feet. 


2         s 

j 

> 

^ 

xl 

•¥ 

Fig.30 

R2: 

=w 

2 

1 

w 


jFig.31 

i 

Fig.  32. 


L 


]w 

Li 


r 


17.  Simple  Beam  with  Concentrated  Load  at  Middle.  —  The 

beam  of  Fig.  30  is  supported  at  the  ends  and  has  a  load  W 
concentrated  at  its  middle. 


34  THE    ELEMENTS  OF  MECHANICS  OF  MATERIALS 

W 
The  reaction  at  each  support  is  —  •     The  bending  moment 

at  any  section  between  the  left  support  and  the  middle  and  dis- 

Wx 
tant  x  from  the  support  is  M  = >  and  increases  directly  as 

x  increases.     At  the  middle,  x  =  —  >  and  M  = At  any  sec- 

2  4 

tion  between  the  middle  and  the  right  support  and  distant  x\ 
from  the  left  support, 

W , 


Wxi      T,7/         ZA      Wx^      TJ7      .  WL 
=  -   ~-W[xi )  =  -   --Wxi-\ = 

2  \  2/  2  2 


M 


and  decreases  as  x  increases,  becoming  o  when  x\  =  L.     The 
maximum  bending  moment  is  therefore  at  the  middle  and  is 


i  .   WL  ™     Wx       ,„     W  ,T 

equal  to  --    Since  the  equations  M  —  -  and  M  =  —  (L  —  Xi) 
4  22 

are  those  of  straight  lines,  the  b  ending-moment  diagram  has  the 
triangular  form  shown  in  Fig.  31. 

W 
Commencing  at  the  left  support  the  shear  is  —  and  remains 

unchanged  until  directly  after  passing  the  middle  section,  when 

W  W 

it  becomes  --  W  =  --  >  changing  sign  at  the  middle  section. 

2  2 

The  shear  diagram  is  then  as  shown  in  Fig.  32. 

18.  Simple  Beam  Uniformly  Loaded.  —  The  beam  of  Fig.  33 
is  the  same  as  that  of  Fig.  30,  but  the  load  W,  instead  of  being 
concentrated  at  the  middle,  is  uniformly  distributed  over  the 
whole  length  of  the  beam  with  w  pounds  per  unit  of  length,  so 
that  wL,  the  whole  load,  is  equal  to  W. 

iv  L 
It  is  evident  that  the  support  reactions  are  each  —  pounds. 

2 

To  find  the  bending  moment  at  any  section  a,  distant  x  from 
the  left  support,  we  may  assume  the  uniform  load  to  be  made 
up  of  a  number  of  parallel  forces  each  equal  to  w.  There  are 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS 


35 


wx  of  these  forces  between  section  a  and  the  left  support,  and 
we  may  substitute  for  them  their  resultant,  wx,  acting  at  their 
center  of  gravity,  which  is  midway  between  the  section  and  the 
support. 


Fig-  35- 


The  bending  moment  at  the  section  a  is  then 


Ma  =  R\x  —  wx  X  - 


wLx      wx2      wx  f  T        ^      wxx' 
=  —  (L-x)=- 


That  is,  the  bending  moment  at  the  section  is  proportional  to 
the  product  of  the  segments  into  which  the  section  divides  the 
beam,  and  the  bending-moment  diagram  is  therefore  a  parabola, 
as  shown  in  Fig.  34,  having  its  axis  vertical  and  under  the  middle 
of  the  beam;  for  it  is  a  property  of  the  parabola  that  if  a  diameter 
be  drawn  to  intersect  a  chord  the  product  of  the  segments  of 
the  chord  is  proportional  to  the  length  of  that  part  of  the  diam- 


36  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

eter  included  between  its  vertex  and  its  point  of  intersection 
with  the  chord. 

The  maximum  bending  moment  is  evidently  at  the  middle  of 

7^J  /"  'V  7^*1* 

the  beam,  so  if  in  the  general  expression  Ma  =  ---  for 

2  2 

the  bending  moment  we  let  x  =  —  >  we  get 

2 


_  WL 

Mmax  -  -  —   -  —  —  -  —  —  > 

4OOO 

in  which  W  —  wL  is  the  whole  weight.  A  comparison  of  this 
result  with  that  obtained  for  the  maximum-bending  moment  in 
Art.  17  discloses  the  fact  that  a  load  concentrated  at  the  middle 
of  a  simple  beam  occasions  a  bending  moment  twice  as  great 
as  that  due  to  the  same  load  uniformly  distributed. 

The  shear  diagram  of  Fig.  35  may  readily  be  constructed. 
Denoting  the  shear  by  F,  we  shall  have  for  any  section  distant 
x  from  the  left  support, 

T/         r>  WL 

V  =  RI  —  wx  =  --  wx, 

2 

which  is  the  equation  of  a  straight  line,  the  origin  being  at  the 

o  «  »  7" 

left  support.  It  is  seen  that  V  has  its  maximum  value,  —  •» 
when  x  =  o,  and  that  it  decreases  as  x  increases  until,  when 
x  =  —  '  it  becomes  o.  For  values  of  x  greater  than  —  the  value 

2  2 

of  V  becomes  negative,  and  when  x  =  L  the  value  of  V  becomes 
—  —  •  The  straight  line  whose  equation  is  V  =  -  -  wx  is 

2  2 

therefore  rs. 

19.  Dangerous  Section.  —  The  maximum  bending  moment  is 
the  controlling  influence  in  the  design  of  beams,  and  the  section 
at  which  it  occurs  is  known  as  the  dangerous  section. 

It  has  been  shown,  Art.  18,  that  in  the  case  of  a  uniformly 
loaded  beam  the  bending-moment  curve  is  parabolic,  and  that 


B  ENDING-MOMENT  AND   SHEAR   DIAGRAMS  37 

therefore  the  ordinate  representing  the  bending  moment  is  a 
continuous  function  of  x.  The  determination  of  the  dangerous 
section  would  then  be  to  find  that  value  of  x  which  would  make 
the  bending  moment  a  maximum.  To  do  this  we  would  place 
the  first  ^-derivative  of  the  bending  moment  equal  to  zero, 
and  the  resulting  value  of  x  would  be  the  abscissa  of  the  dan- 
gerous section. 

For  example,  the  general  expression  for  the  bending  moment 
of  the  uniformly  loaded  beam  of  Fig.  33  is 

,,       „ 
M  =  RIX 

2 

Then  -  =  RI  —  wx. 

dx 

RI      wL      L 
whence    x  —  -  -  =  —  • 

W          2W          2 

The  maximum  bending  moment  in  this  case,  and  therefore  the 
dangerous  section,  is  at  the  middle  of  the  beam. 

tint  I 

If,  in  the  general  expression wx  for  the  shear  in  Art.  18, 

2 

we  substitute  for  x  the  value  — ,  the  result  is  zero,  from  which 

2 

we  infer  that  in  the  case  of  uniform  loading  the  maximum  bend- 
ing moment  occurs  at  the  section  where  the  shear  is  zero. 

In  the  equation——  =  RI  —  wx  it  will  be  observed  that  the 

dx 

second  member  is  the  general  expression  for  the  shear  over  the 
whole  beam,  from  which  we  may  also  infer  that  the  first  deriva- 
tive of  the  bending  moment  is  the  shear. 

If,  however,  in  addition  to  the  uniformly  distributed  load  the 
beam  were  subjected  to  one  or  more  concentrated  loads,  or  if 
there  were  no  uniform  load  and  the  beam  were  subjected  simply 
to  one  or  more  concentrated  loads,  the  ordinate  representing  the 
bending  moment  would  no  longer  be  a  continuous  function  of 


38  THE   ELEMENTS  OF  MECHANICS  OF   MATERIALS 

x.  The  first  derivative  of  the  bending  moment  at  any  section 
would,  however,  still  be  the  shear  at  that  section;  but  the  shear 
at  the  dangerous  section  might  not  be  zero,  because  the  shear 
is  not  necessarily  zero  at  any  section.  If  the  construction  of 
the  shear  diagram  shows  that  the  shear  is  not  zero  at  any  sec- 
tion, it  will  also  show  that  at  some  one  or  more  sections  the 
shear  suddenly  changed  sign,  or  passed  through  zero,  and  the 
maximum  bending  moment  will  be  found  to  occur  at  one  of 
these  sections. 

20.  Simple  Beam  with  two  Equal  and  Symmetrically  Placed 
Loads.  —  The  beam  of  Fig.  36  is  supported  at  the  ends  and  has 
two  equal  and  symmetrically  placed  loads. 


At  the  section  under  the  load  nearest  the  left  support,  M  — 
Wa\  at  any  section  between  the  loads  and  distant  x  from  the  left 
support  the  bending  moment  is  Mx  =  Wx  —  W  (x  —  a)  =  Wa ; 
at  any  section  between  the  second  load  and  the  right  support 


BENDING-MOMENT  AND    SHEAR  DIAGRAMS  39 

and  distant  Xi  from  the  left  support  the  bending  moment  is 
MX1=  Wx1  -  W  (xi  -  a)  -  W[xi  -(L-  a)}  =  WL-  W*i.  For 
Xi  =  L  in  the  last  equation  we  get  MXl=o.  From  these  expres- 
sions for  the  bending  moments  the  diagram  of  Fig.  37  was  con- 
structed. 

Commencing  at  the  left  support  the  shear  is  W  and  is  unchanged 
until  the  first  load  is  passed,  when  it  becomes  W  —  W  =  o. 
There  is  no  further  change  until  the  second  load  is  passed,  when 
the  shear  becomes  W  —  W  —  W  =  —  W,  and  so  continues  to  the 
right  support.  The  shear  diagram  is,  then,  as  shown  in  Fig.  38. 

21.  B ending-Moment  and  Shear  Diagrams  of  Cantilevers.  - 
The  cantilever  of  Fig.  39  has  a  concentrated  load  W  at  the  end, 
a  concentrated  load  W\  at  an  intermediate  point  between  the 
end  and  the  support,  and  a  uniformly  distributed  load  of  w 
pounds  per  foot  over  a  portion  of  its  length.  The  construction 
of  the  b ending-moment  and  shear  diagrams  of  this  beam  will 
serve  to  illustrate  the  three  cases  of  a  cantilever:  (a)  Loaded  at 
the  end.  (b)  A  concentrated  load  at  some  point  between  the 
end  and  the  support,  (c)  Loaded  uniformly  with  w  pounds  per 
unit  of  length. 

From  our  definition  of  a  bending  moment  it  will  be  seen  that 
the  bending  moment  is  o  at  the  free  end  and  a  maximum  at  the 
wall;  and  the  tendency  being  to  bend  the  beam  concave  down- 
ward all  the  bending  moments  are  negative. 

In  the  construction  of  the  bending-moment  diagram  of  Fig.  40 
the  following  data  were  used 

Linear  scale,  0.2  inch  =  i  foot;  bending-moment  scale,  i  inch 
in  depth  =  800  pounds-feet.  L  =  10  feet  =  2  inches  to  scale; 
7-1  =  7  feet  =  I-4  inches  to  scale;  r2  =  4  feet  =  0.8  inch  to  scale; 
W  =  32  pounds;  Wi  =  44  pounds;  w  =  48  pounds  per  foot  run. 

If  the  beam  were  loaded  only  with  W  at  the  extremity  the 
bending  moment  at  the  wall  would  be 
—  WL  =  -32X10  =  -3  20  lbs;-ft.  =  —  §£$  =  —  0.4  inch  to  scale, 


40  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

which  would  be  the  length  of  the  ordinate  no  at  the  wall  repre- 
senting the  bending  moment  due  to  Wj  and  mno  would  be  the 
complete  bending-moment  diagram. 


Fig.  41. 

The  addition  of  the  concentrated  load  Wi  would  occasion  an 
additional  bending  moment  at  the  wall  of 
—  WiTi  =  —  44  X  7  =  —  308  =  —  fat  =  —  0.385  inch  to  scale, 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  41 

which  would  be  the  length  of  the  ordinate  np  at  the  wall  rep- 
resenting the  bending  moment  due  to  W\.  Drawing  the  line  sp 
we  would  then  have  mspo  as  the  complete  bending-moment 
diagram  due  to  W  and  W\. 

The  addition  of  the  uniform  load  occasions  a  bending  moment 
at  the  wall  of 

—  wr2X-  =  —  48  X  4  X  2  =  —  384  pounds-feet 
=  —  t  <m  =  —  0.48  inch  to  scale, 

which  would  be  the  length  of  the  ordinate  ot  at  the  wall  repre- 
senting the  bending  moment  due  to  the  uniform  load,  taking 
mo  as  a  base  line.  The  locus  of  the  extremities  of  the  ordinates 
representing  the  bending  moments  at  all  other  sections  of  the 
uniformly  loaded  part  would  be  the  parabola  having  its  apex 
at  v  and  passing  through  /. 

The  complete  bending-moment  diagram  for  all  the  loads  is 
therefore  msptv.  This  can  be  checked  by  finding  at  once  the 
bending  moment  at  the  wall  due  to  all  the  loads,  thus: 

M=  -WL-  Wiri-  w2  X  -  =  -  32  X  10  -  44  X  7  -  48  X  4  X  2 

2 


=  1012  pounds-feet  =  VW  =  1-265  inches  to  scale. 

The  part  of  the  bending-moment  diagram  due  to  the  uniform 
load  being  negative  properly  belongs  below  the  base  line  mo,  as 
shown  by  the  dotted  lines,  but  it  has  been  placed  above  mo  in 
order  to  obtain  a  solid  diagram  from  which  the  bending  moment 
at  any  section  under  the  uniform  load  may  be  obtained  by  one 
measurement. 

In  the  construction  of  the  shear  diagram  of  Fig.  41  the  load 
scale  was  taken  as  i  inch  =  160  pounds. 

If  W  were  the  only  load  on  the  beam  the  shear  throughout  the 
beam  would  be  -  W  =  -32  pounds  =  -  f&  =0.2  inch  to  scale. 
Between  the  point  of  application  of  Wi  and  the  wall  the  shear  is 
increased  by  the  amount  -  W\  =  -  44  pounds  =  -  A%  =  -  0.275 


42  THE    ELEMENTS  OF  MECHANICS  OF   MATERIALS 

inch  to  scale.  The  addition  of  the  uniform  load  increases  the 
shear  uniformly  from  zero  at  its  commencement  to  —  wr^  pounds 
=  —  48  X  4  =  —  192  pounds  =  —  !!§  =  —  1.2  inches  to  scale  at 
the  wall. 

This  can  be  checked  by  finding  at  once  the  shear  at  the  wall 
due  to  all  the  loads,  thus: 

Shear  at  wall  =  —  W  —  W\  —  wr%  =  — 32—  44  —  48X4  =  —  268  Ibs. 
=  —  fit  =  —  1-675  inches  to  scale. 

22.  Beam  with  Overhanging  Ends  and  Uniformly  Loaded.  - 
The  beam  of  Fig.  42  is  uniformly  loaded  with  w  pounds  per  unit 
of  length  and  overhangs  the  supports  a  distance  a  at  each  end. 
Consider,  quite  independently  of  each  other,  the  two  overhangs 
as  cantilevers  and  the  part  between  the  supports  as  a  simple 
beam. 

Since  the  overhangs  tend  to  bend  concave  downward  the 
bending  moments  due  to  them  are  negative.  The  bending 
moment  at  each  support  due  to  the  overhanging  cantilevers  is 

9 

—  wa  X  -  = and,  as  we  have  seen  in  Art.  21,  the  para- 

2  2       ' 

bolic  curves  joining  the  extremities  of  the  beam  with  the  extremi- 
ties of  the  ordinates  representing  the  bending  moments  at  the 
supports  form  the  bending-moment  diagrams  of  the  overhangs, 
Fig.  43.  Disregarding  the  load  on  the  central  span  of  the  beam, 
the  reactions  at  the  supports  are  each  wa.  Then  the  bending 
moment  for  any  section  between  the  supports  and  distant  x  from 
the  left  support  is 

•K/T  !a      I         \  Wa2 

M  =  wax  —  w  a  -  +  x\  = > 

\2  /  2 

showing  the  bending  moment  due  to  the  overhanging  loads  to 
be  constant  between  the  supports  and  negative  in  character. 
Therefore  the  negative  part  below  the  base  line  mn  of  Fig.  43 
represents  the  bending  moments  of  the  whole  beam  due  to  the 
overhangs. 


BEND  ING-MOMENT   AND    SHEAR   DIAGRAMS 


43 


Fig.  45- 


44  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

The  uniformly  loaded  central  span  of  length  b  occasions  a 
reaction  at  each  support  of  —  ,  and  as  the  tendency  is  to  bend 

the  beam  concave  upward  the  bending  moments  are  positive. 
The  bending  moment  at  any  section  between  the  supports  and 
distant  x  from  the  left  support  is 

T.,      wbx      wx2      wx  /t        N 
M  =  -  --  =  —  (b  -  x), 

2  2  2 


which,  as  we  have  seen,  is  the  equation  of  a  parabola  having  its 
axis  vertical  over  the  middle  of  the 
for  the  maximum  bending  moment 


axis  vertical  over  the  middle  of  the  beam.     When  x  =  -  we  have 

2 


The  parabola  above  the  base  line  of  Fig.  43  is  the  bending-moment 
diagram  of  the  part  of  the  beam  between  the  supports  due  alone 
to  its  load,  and  is  positive.  By  superposition  the  resultant 
bending-moment  diagram  of  the  whole  beam,  Fig.  44,  is 
obtained. 

Considering  the  beam  in  its  entirety,  the  shear  of  the  overhang 
at  the  left  end  increases  uniformly  from  zero  at  the  extremity  to 
—  wa  at  the  left  support.  The  reactions  due  to  the  load  are 

each  wa  +  —  »  so  that  at  the  instant  of  passing  the  left  support 

the  shear  becomes 

.  wb      wb 
—  wa  +  wa  H  --  =  —  • 

2  2 

At  the  middle  the  shear  becomes 

.  wb      wb 

—  wa  -\-wa  -\  ----  =  o. 
2         2 

Immediately  to  the  left  of  the  right  support  the  shear  is 

.  wb  wb 

—  wa  +  wa  H  ---  wb  =  —  — 
2  2 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  45 

Just  passing  the  right  support  the  shear  is 

.  wb        ,    .  wb 

—  wa  +  wa  H z^o  +  wa  +  —  =  wa. 

2  2 

At  the  extreme  right  end  the  shear  is 

.  wb         ,                wb 
—  wa  +  wa  H wo  -{-wa  -\ wa  =  o. 

2  2 

These  values  of  the  shear  enable  the  shear  diagram  of  Fig.  45  to 
be  constructed. 

23.  Simple  Beam  Loaded  Uniformly  over  a  Part  of  its  Length 
Adjoining  one  Support.  —  The  beam  of  Fig.  46  is  supported  at 
the  ends  and  loaded  with  w  pounds  per  unit  of  length  for  a  dis- 
tance r  from  the  left  support. 

To  find  the  support  reactions  we  take  moments  about  the 
supports,  thus: 

=  wr(  L  —  -J>     whence     RI  =  wr(i r—j\ 

T  r>          wr*  T-  T>  wr<1 

LR%  =  — >     whence     R^  = 

2  2  L 

The  general  expression  for  the  bending  moment  within  the 
limits  of  the  loaded  part  of  the  beam  is 

,,       „         wx2 
M  =  R\x 

2 

For  x  =  r  we  get 

K/T        r>  wr<i       wr*f          r\ 

M  =  Rir =  —   i  -— ) 

2  2     \  LI 

as  the  bending  moment  at  the  section  where  the  load  ceases. 

On  the  base  line  ab}  Fig.  47,  erect  the  ordinate  cd  under  the 
section  of  the  beam  at  the  right  extremity  of  the  loaded  part, 

2  /  \ 

and  make  it  equal  in  length  to  —  ( i  — r)  to  a  chosen  bending- 

2     \  J~tf 

moment  scale. 


THE    ELEMENTS   OF  MECHANICS   OF   MATERIALS 


The  general  expression  for  the  bending  moment  at  any  section 
between  the  loaded  part  and  the  right  support  is 

IT  r>  /  ?\         Wr*  I  Xi\ 

M  =  R&i  -wrlxi  --)  =  • —   i  -  -M, 

\  2/  2     V  LI 


w  Ibs.per  ft. 

^                             i 
J 

:                          I 

i  •  1 

'  Ri  J 

IV                      "~*1 

!/ 

Xl                                   ;| 

Fig.  46                                   | 

Fig.  48 


the  equation  of  a  straight  line  which  gives  a  zero  value  for  M 
when  Xi  =  L.  The  triangle  bed,  therefore,  is  the  diagram  of  the 
bending  moments  for  the  unloaded  part  of  the  beam. 


BENDING-MOMENT   AND    SHEAR  DIAGRAMS  47 

The  maximum  bending  moment  will  evidently  occur  at  some 
section  within  the  limits  of  the  load.  To  locate  this  section  we 
proceed  thus: 

Within  the  limits  of  the  load 


dM 

—  =  RI  -  wx, 
dx 

-p 

RI  —  wx  =0,    whence   x  =  —  • 

w 

The  maximum  bending  moment  therefore  occurs  at  the  section 
-p 

distant  —  -1  from  the  left  support.     Substituting  this  value  of  x 

w 

in  the  general  expression  for  the  bending  moment  within  the 
limits  of  the  load  we  get 

_  RI 

•Ml.  max  — 


<> 

W  2  W  2  W 

-p 

Under  the  section  distant  —  from  the  left  support  erect  the 

w 

7?  2 

ordinate  ef  and  make  it  equal  in  length  to  —  -  to  the  chosen 

2  w 

scale.  This  ordinate  is  the  axis  of  the  parabola  representing 
the  bending  moments  due  to  the  uniform  load,  and  since  this 
parabola  must  pass  through  the  points  a  and  d  it  is  readily  con- 
structed as  shown,  and  the  bending-moment  diagram  of  the 
beam  completed. 

The  general  expression  for  the  shear  over  the  loaded  part  of 
the  beam  is  RI  —  wx,  which,  for  x  =  o,  gives  Ri  as  the  shear 
at  the  left  support.  For  x  —  r  the  shear  becomes  RI  —  wr 

=  wrli  --  -  )  —  wr  =  —  —  -  =  —  RZ   and  remains  unchanged 
\        2  L]  2  L 

over  the  unloaded  part  of  the  beam.  From  the  shear  values  just 
found  the  shear  diagram  of  Fig.  48  is  constructed,  and  it  will  be 


48  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

observed  that  the  maximum  bending  moment  occurs  at  the 
section  where  the  shear  is  zero. 


<          a             > 

r                     ''                      » 

> 

w  Ibs.per  ft. 

^ 

<  X^  H 
1 

1 

Fig.  49 

24.  Simple  Beam  Loaded  Uniformly  over  a  Part  of  its  Length 
not  Adjoining  Either  Support.  —  The  beam  of  Fig.  49  is  sup- 
ported at  the  ends  and  loaded  with  w  pounds  per  foot  for  a 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  49 

distance  r  commencing  at  a  distance  a  from  the  left  support  and 
terminating  at  a  distance  b  from  the  right  support. 

To  find  the  support  reactions  we  take  moments  about  the 
supports,  thus: 

whence    R^ 


LRZ  =  wrt  a  +  -)>    whence    R2  =  ~(0  +  -V 

V  2/  L  V  2/ 

The  general  expression  for  the  bending  moment  between  the 
left  support  and  the  section  where  the  loading  begins  is 


the  equation  of  a  straight  line.  For  xi  =  o,  M  =  o;  for  #1  =  a, 
M  =  —r(b  +  -]•  On  the  base  line  mn,  Fig.  50,  and  under  the 
section  distant  a  from  the  left  support,  erect  the  ordinate  cd 
and  make  it  equal  in  length  to  ^y^  (  b  +  -j  to  some  chosen  scale 

for  bending  moments.  Join  m  and  d  and  the  triangle  mcd  is 
then  the  diagram  of  the  bending  moments  of  the  unloaded  part 
of  the  beam  adjoining  the  left  support. 

The  general  expression  for  the  bending  moment  of  the  un- 
loaded part  of  the  beam  adjoining  the  right  support  is,  taking 
moments  to  the  right, 


the  equation  of  a  straight  line.     For  Xz  =  a  +  r, 


and  for  #2  =  o,  M  =  o. 
At  a  distance  of  a  +  r  from  the  left  support  erect  the  ordinate 

gh  and  make  it  equal  in  length  to  wr  I  a  -\--\li  --  -  —  V     It 


50  THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 

will  represent  the  bending  moment  at  the  section  under  the  right 
extremity  of  the  load.  Join  h  and  n  and  the  triangle  ngh  is  the 
diagram  of  the  bending  moments  of  the  unloaded  part  of  the 
beam  adjoining  the  right  support. 

The  maximum  bending  moment  evidently  occurs  at  some 
section  within  the  limits  of  the  loaded  part  of  the  beam.  To 
locate  this  section  we  proceed  thus  : 

Within  the  limits  of  the  load 

M  =  Rix*--  (x,-a)\ 

2 

dM 

—  -  =  Ri  -w(x3  -a), 
dx3 

-p 

RI  —  w  (x3  —  a)  =  o,    whence    x3  =  —  -  +  a. 

w 

The  maximum  bending  moment  occurs,  therefore,  at  the  section 

•p 
distant  —  -  +  a  from  the  left  support.     Substituting  this  value 

of  x3  in  the  general  expression  for  the  bending  moment  within 
the  limits  of  the  load  we  get 


W 

r> 

Under  the  section  distant  —  -  +  a  from  the  left  support  erect 

w 

the  ordinate  ef  and  make  it  equal  to  RI(  —  -  -f-  a]  to  the  chosen 

\2W          I 

scale.  This  ordinate  is  the  axis  of  the  parabola  representing 
the  bending  moments  due  to  the  uniform  load,  and  since  this 
parabola  must  pass  through  d  and  h  it  is  readily  constructed  as 
shown,  and  the  bending-moment  diagram  of  the  whole  beam 
completed. 

The  shear  over  the  unloaded  part  of  the  beam  adjoining 
the  left  support  is  Rit  and  over  the  loaded  part  of  the  beam 
it  is  RI  —  w  (x3  —  a),  which,  for  #3  =  a  +  r,  gives  for  the 


BENDING-MOMENT  AND    SHEAR   DIAGRAMS  51 

shear  at  the  section  at  the  right  extremity  of  the  load,  RI  —  wr 

wr/,    ,    r\  wr/,    .    r       T\          wr  /     .  r\          D 

=  —  (b  +  -   -  wr  =  —- (b  H L    =-—   a  +-    =-£2,and 

L\        21  L\        2         /          L\2/ 

continues  unchanged  to  the  right  support.  From  these  shear 
values  the  shear  diagram  of  Fig.  51  is  constructed,  and  it  will 
be  observed  that  the  maximum  bending  moment  occurs  at  the 
section  where  the  shear  is  zero. 

EXAMPLE.  —  The  beam  of  Fig.  52  is  32  feet  long  and  overhangs 
the  right  support  6  feet  and  the  left  support  8  feet.  In  addition 
to  a  uniformly  distributed  load  of  20  pounds  per  foot  it  carries 
concentrated  loads  of  200  pounds  and  390  pounds  as  shown. 
Find  the  maximum  positive  and  negative  bending  moments 
and  construct  the  bending-moment  and  shear  diagrams. 

Solution.  —  The  support  reactions  must  first  be  found  by 
taking  moments  about  the  supports;  thus, 

18  RI  +  20  X  6  X  3  =  20  X  26  X  13  +  200  X  22  +  390  X  6, 
whence  RI  =  730  pounds. 

18  R2  +  20  X  8  X  4  +  200  X  4  =  20  X  24  X  12  +  390  X  12, 
whence  R2  =  500  pounds 

In  complicated  problems  like  this  it  is  the  best  practice  to 
calculate  the  bending  moments  at  various  sections  of  the  beam 
and  use  the  results,  when'  reduced  to  scale,  as  the  lengths  of 
ordinates  in  the  construction  of  the  bending-moment  diagram. 
In  this  instance  the  linear  scale  will  be  taken  as  o.  i  inch  =  i  foot, 
and  the  bending-moment  scale  as  i  inch  =  1000  pounds-feet. 

For  the  purpose  of  constructing  the  bending-moment  diagram 
these  bending  moments  are  obtained  by  calculation 

At  the  section  under  Wi 

M  —  —  20X4X2=—  1 60  pounds-feet. 

At  the  left  support 

M  =—  20X8X4  —  200  X  4  =  —  1440  pounds-feet. 


52  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  53 

At  the  section  2  feet  to  the  right  of  the  left  support 

M  =  730  X  2  —  20X10X5  —  200  X  6  =  —  740  Ibs.-ft. 

At  the  section  4  feet  to  the  right  of  the  left  support 

M  =  730  X  4  -  20  X  12  X  6  -  200  X  8  =  -  i2olbs.-ft. 

At  the  section  6  feet  to  the  right  of  the  left  support 

M  —  730  X6  —  20X14X7  —  200  X  10  =  420  Ibs.-ft. 

At  the  section  10  feet  to  the  right  of  the  left  support 

M  =  730  X  10  —  20  X  18  X  9  —  200  X  14  =  1 260  Ibs.-ft. 

At  the  section  under  Wz 

M  =  730  X  12  —  20  X  20  X  10  —  200  X  16  =  1560  Ib.-ft. 

At  the  section  4  feet  to  the  left  of  the  right  support 

M  =  730  X  14  —  20  X  22  X  ii  —  200  X  18  —  390  X  2 
=  1000  pounds-feet. 

At  the  right  support 

M  =  730  X  18  —  20  X  26  X  13  —  200  X  22  —  390  X  6 
=  —  360  pounds-feet. 

At  the  right  end 

M  =  730  X  24  —  20  X  32  X  16  —  200  X  28  —  390  X  12 
+  500  X  6  =  o. 

Reducing  these  bending-moment  results  to  scale  by  dividing 
each  by  1000,  and  using  the  results  as  the  lengths  of  ordinates 
with  respect  to  mn  as  an  axis,  the  bending-moment  diagram  of 
Fig.  53  was  constructed. 

The  maximum  positive  bending  moment  is  seen  to  be  at  the 
section  under  W^  the  ordinate  measuring  1.56  inches,  which, 
reduced  to  scale,  represents  1560  pounds-feet.  The  maximum 
negative  bending  moment  is  under  the  left  support  and  measures 
1440  pounds-feet.  The  bending  moment  which  is  numerically 
greatest,  whether  positive  or  negative,  is  the  one  to  be  considered 
in  the  design  of  the  beam. 


54  THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 

The  shear  diagram  of  Fig.  54  may  be  obtained  by  taking  the 
algebraic  sum  of  the  forces  to  the  left  of  different  sections,  thus: 

Commencing  at  the  left  end,  the  shear  increases  uniformly  and 
negatively  at  the  rate  of  20  pounds  per  foot  from  the  value  zero 
until,  at  the  section  immediately  to  the  left  of  W\,  it  becomes 

—  20  X  4  =  —  80  pounds, 

which,  to  a  scale  of  800  pounds  to  the  inch,  is  represented  in 
the  diagram  of  Fig.  54  by  —  o.i  inch. 
Just  passing  W\  the  shear  becomes 

—  80  —  200  =  —  280  pounds, 

which,  to  scale,  is  represented  by  —  0.35  inch. 

At  the  section  just  to  the  left  of  the  left  support  the  shear 
becomes 

-i  20  X  8  —  200  =  —  360  pounds, 

which,  to  scale,  is  represented  by  —  0.45  inch. 
Just  passing  the  left  support  the  shear  becomes 

—  20  X  8  —  200  +  730  =  370  pounds, 

which,  to  scale,  is  represented  by  0.4625  inch;  and  so  on  to  the 
right  end  of  the  beam. 

The  shears  at  the  different  sections  may  be  obtained  also  by 
taking  the  first  derivatives  of  the  bending  moments  at  the 
sections,  thus: 

Reckoning  from  the  left  end  of  the  beam,  the  expression  for 

20  X2 

the  bending  moment  for  any  section  up  to  W\  is  — 


2 


Then  —  =—  2ox  =  —  20X4=  —  80  pounds 

dx 

for  the  shear  just  to  the  left  of  W\. 

For  any  section  between  W\  and  the  left  support 

..,  20  X2  (  N 

M  = —  200  (x  —  4). 

2 


BENDING-MOMENT  AND    SHEAR   DIAGRAMS  55 

Then  -  =  —  20  x  —  200  =  —  20  X  4  —  200  =  —  280  pounds 

ax 

for  the  shear  at  the  section  just  to  the  right  of  Wi\  and 

=  —  20  x  —  200  =  —  20  X  8  —  200  =  —  360  pounds 

doc 

for  the  shear  at  the  section  just  to  the  left  of  the  left  support. 
For  any  section  between  the  left  support  and  W2 

M  =  Rl  (x  -  8)  -  Wi  (x  -  4)  -  — 

2 

Then     — —  =  RI  —  Wi  —  2ox  =  730  —  200  —  20  X  8  =  370  Ibs. 

doc 

for  the  shear  at  the  section  just  to  the  right  of  the  left  support; 
and  so  on  to  the  right  end  of  the  beam. 

The  completion  of  the  shear  diagram  by  either  method  is  left 
as  a  study  for  the  student. 

25.  Moving  Loads.  —  A  load  moving  over  a  structure,  such 
as  a  train  moving  over  a  bridge,  is  a  live  load,  sometimes  known 
as  a  rolling  load.  A  study  of  the  bending  moment  and  shear 
effects  of  the  different  conditions  of  loads  moving  over  structures 
is  beyond  the  scope  of  this  work,  and  only  the  simple  case  of  a 
single  concentrated  load  moving  over  a  beam  will  here  be  given. 
For  an  extended  study  of  moving  loads  recourse  must  be  had 
to  a  more  advanced  treatise  on  the  Mechanics  of  Materials. 

26  Simple  Beam  with  a  Concentrated  Moving  Load.  —  Ne- 
glecting the  weight  of  the  beam  of  Fig.  55,  suppose  the  load  W 

to  move  from  left  to  right. 

W  (L  —  x] 
By  moments  about  the  right  support  we  find  RI  =  — ^— • 

J_j 

While  the  load  is  between  the  left  support  and  some  section  a 
distant  r  from  the  support,  the  bending  moment  is 

M  =  Rir  -  W  (r  -  x)  =  Wr  (L~  x)  -  W(r-x)  = 

This  result  must  be  positive,  since  r  is  less  than  L,  and  will 


50  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

increase  in  value  as  x  increases,  until  the  load  reaches  a,  where  r 
becomes  x,  and  we  shall  have  for  the  bending  moment  at  the  load, 

M  =  Wx  _W£.  (3) 

Li 


When  the  load  passes  section  a  and  moves  toward  the  right 
support,  the  bending  moment  at  a  becomes  Rtf  and  decreases 
in  value,  since  RI  becomes  less  as  W  approaches  the  right  sup- 
port. When  W  reaches  the  right  support,  RI  becomes  zero 
and  the  bending  moment  at  a  becomes  zero.  The  bending 


BENDING-MOMENT  AND   SHEAR   DIAGRAMS  57 

moment  at  a  is  greatest  when  the  load  is  at  #,  and  therefore  the 
bending  moment  at  any  section  is  a  maximum  when  the  load  is 
at  the  section. 

To  find  the  dangerous  section,  the  value  of  x  in  equation  (3) 
which  makes  M  a  maximum  must  be  found.     Thus, 


dM      TJ/      2Wx  L 

——  =  W  --  —  =  o;     whence,     x  =  —  - 
ax  L  2 

The_maximum  bending  moment  at  the  load,  and  therefore  at 
the  dangerous  section,  is  then  at  the  middle  of  the  beam. 

Since  the  maximum  bending  moment  at  any  section  occurs 
when  the  load  is  at  the  section,  the  diagram  of  maximum  bend- 
ing moments,  Fig.  56,  is  constructed  from  equation  (3),  which  is 

that  of  a  parabola,  the  maximum  ordinate  -  being  at   the 

4 

middle,  where  x  =  —  • 

2 

The  shear  diagram  is  shown  in  Fig.  57.  At  the  instant  the 
load  starts  to  move  to  the  right  from  the  left  support,  RI  =  W 
and  R%  =  o.  As  the  movement  continues  the  shear  V  to  the 
left  of  the  load  is 


a  straight-line  equation.  At  the  left  support  x  =  o  and  V  =  W', 
Sit  the  right  support  x  =  L  and  V  =  o.  The  line  of  equation 
(4)  is  then  bd,  and  the  diagram  for  the  shear  to  the  left  of  the 
load  takes  the  triangular  form  bed. 

To  the  right  of  the  load  during  the  movement  the  shear  is 

Wv 

V  =  Rl-W  =  -R,=-^,  (5) 

J_j 

a  straight-line  equation.  At  the  left  support  x  =  o  and  V  =  o; 
at  the  right  support  x  =  L  and  V  =  —  W.  The  line  of  equa- 
tion (5)  is  then  ce,  and  the  diagram  for  the  shear  to  the  right  of 
the  load  takes  the  triangular  form  ced. 


THE    ELEMENTS   OF   MECHANICS  OF   MATERIALS 


27.  B ending-moment  Influence  Line.  —  An  ordinate  of  the 
bending-moment  diagram  of  Fig.  56  represents  the  greatest  bend- 
ing moment  produced  by  the  rolling  load  at  the  section  corre- 
sponding to  the  ordinate,  but  it  is  obvious  that  as  the  load 
moves  over  the  beam  the  bending  moment  at  the  section  changes 
continuously.  The  graphic  representation  of  the  variation  of 
the  bending  moment  due  to  a  rolling  load  at  a  given  section  of  a 
beam  or  structure  is  called  the  influence  line  of  the  bending 
moment  at  the  section. 


< 

c 

w 
)    5 

h 

I 

Fig.  58 

Let  x  denote  the  distance  from  the  left  support  of  a  con- 
centrated load  W  moving  over  the  beam  of  Fig.  58,  and  let 


BENDING-MOMENT  AND   SHEAR  DIAGRAMS  59 

r   denote    the  distance  of    a    given    section   a   from  the  left 
support. 
While  W  is  at  the  right  of  a  we  shall  have 


If  W  be  taken  as  the  unit  load  of  one  pound  or  of  one  ton  we 
shall  have 


(6) 


the  equation  of  the  straight  line  mnt  Fig.  59,  the  locus  of  all 
values  of  Ma  between  the  limits  of  x  =  L  and  x  =  r. 
When  the  load  moves  to  the  left  of  a  we  have 


For  the  unit  load  we  shall  then  have 

*.-**¥>•  (7) 

Equation  (7)  is  that  of  the  straight  line  on,  which  is  the  locus 
of  the  values  of  Ma  between  the  limits  x  =  r  and  x  =  o. 

For  x  =  r  in  equations  (6)  and  (7)  we  have  the  same  value, 

Ma  =  -  —  —  -  i    hence  the  two  lines  intersect  at  n,  and  mno  is 
]-j 

the  influence  line  of  the  bending  moment  at  section  a.  The 
ordinate  to  the  influence  line  under  any  section  measures  the 
bending  moment  at  a  due  to  the  unit  load  when  moving  over 
the  section,  and  the  whole  moment  is  found  by  multiplying  the 
ordinate  by  W.  Thus,  the  bending  moment  at  section  a  when 
the  load  passes  over  section  b  is  Wy. 

When  the  load  comes  on  the  beam  from  the  right,  the  bending 
moment  at  a  increases  uniformly  from  zero  to  the  maximum  at 
w,  and  then  decreases  uniformly  to  the  value  zero  at  the  left 
support. 


60  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

28.  Shear  Influence  Line.  —  The  influence  line  for  the  shear 
at  section  a  of  Fig.  58  may  be  expressed  graphically. 

While  the  moving  load  W  is  to  the  right  of  a  the  shear  V 
Sit  a  is 


.  and  for  the  unit 


a 

-L/ 

Equation  (8)  is  that  of  the  straight  line  m'n'  ',  Fig.  60,  the  locus 
of  the  shear  between  the  limits  x  =  L  and  x  =  r. 
When  the  load  moves  to  the  left  of  a  the  shear  is 


<r  v 

Va  =  Rl-W  =  -R2  =  -  •!-—;  and  for  the  unit  load,F0=  —  f  •    (9) 

jLj  JLs 

Equation  (9)  is  that  of  the  straight  line  which  is  the  locus  of 
Va  between  the  limits  x  =  r  and  x  =  o. 

The  slopes  of  equations  (8)  and  (9)  being  the  same,  the  lines 
are  parallel,  and  o'n"  parallel  to  n'm'  is  the  straight  line  of 
equation  (9),  and  the  broken  line  m'n'n"o'  is  the  influence  line 
of  the  shear  at  the  section  a  while  the  load  moves  over  the  beam. 

When  the  unit  load  comes  on  the  beam  from  the  right  the 
shear  at  a  increases  uniformly  from  zero  to  the  value  snr  at  the 
section  immediately  to  the  right  of  a.  As  the  load  passes  a 
the  shear  suddenly  drops  by  the  amount  of  the  unit  load  and 
becomes  negative,  and  then  uniformly  decreases  numerically 
until  it  becomes  zero  at  the  left  support. 

The  ordinate  yf  measures  the  shear  at  section  a  when  the  unit 
load  passes  over  section  6,  and  the  shear  at  a  due  to  the  whole 

load  is  Wyf. 

PROBLEMS 

1.  A  uniform  beam  25  feet  in  length,  whose  weight  is  disregarded,  is 
supported  at  the  ends  and  has  a  concentrated  load  of  400  pounds  at  9  feet 
from  the  left  support,  and  one  of  500  pounds  at  18  feet  from  the  left  sup- 
port.    Construct  the  bending-moment  and  shear  diagrams. 

2.  Construct  the  bending-moment  and  shear  diagrams  of  the  beam  of 
problem   i,  taking  into   consideration  the  weight  of  the  beam,  which  is 
500  pounds. 


SHEAR   DIAGRAMS  6l 

3.  A  beam  12  feet  long  is  supported  at  the  ends  and  loaded  with  a 
weight  of  3  tons  at  a  point  2  feet  from  one  end.    Find  the  bending  moment 
and  shear  at  the  middle  of  the  beam.  Ans.   3  tons-ft;  0.5  ton. 

4.  A  cantilever  projects  10  feet  from  a  wall  and  carries  a  uniform  load 
of  60  pounds  per  foot;  it  also  supports  three  concentrated  loads  of  100 
300,  and  500  pounds  at  distances  from  the  wall  of  2,  5,  and  9  feet  respec- 
tively.    Construct  the  bending-moment  and  shear  diagrams. 

5.  A  beam  overhangs  both  supports  equally,  carries  a  uniform  load  of 
80  pounds  per  foot,  and  has  a  load  of  1000  pounds  in  the  middle.     The 
length  of  the  beam  is  15  feet,  and  the  distance  between  the  supports  8 
feet.     Construct  the  bending-moment  and  shear  diagrams. 

6.  A  beam  with  equal  overhanging  ends  is  loaded  with  three  equal  con- 
centrated loads,  one  at  each  end  and  one  at  the  middle.     Find  the  distance 
between  the  supports,  in  terms  of  the  total  length  L,  when  the  bending 
moment  at  the  middle  is  equal  to  the  bending  moment  over  the  supports. 
At  what  sections  is  the  bending  moment  zero  ? 

Ans.  --  ;    -from  each  support. 

7.  A  beam  32  feet  long  and  supported  at  the  ends  is  loaded  uniformly 
for  a  distance  of  20  feet  from  the  left  support  with  0.8  ton  per  foot.     Lo- 
cate the  dangerous  section,  and  find  by  calculation  the  maximum  bending 
moment  and  the  bending  moments  at  the  middle  of  the  load,  at  the  end  of 
the  load,  and  at  the  section  5  feet  from  the  right  support.     Construct  the 
bending-moment  and  shear  diagrams. 

Ans.  13.75  ft.  from  left  support;  75.625  tons-ft.;  70  tons-ft. •  60  tons-ft.; 
25  tons-ft. 

8.  A  beam  30  feet  long  and  supported  at  the  ends  is  loaded  uniformly 
with  0.4  ton  per  foot  for  a  distance  of  10  feet,  the  load  commencing  at  a 
distance  of  8  feet  from  the  left  support  and  terminating  at  the  section  12 
feet  from  the  right  support.     Locate  the  dangerous  section,  and  find  by 
calculation  the  maximum  bending  moment  and  the  bending  moments  at 
the  extremities  of  the  load.     Construct   the  bending-moment  and  shear 
diagrams. 

Ans.   13.667  ft.  from  the  left  support;  24.56  tons-ft.;  20.8  tons-ft.;  18.14 
tons-ft. 


CHAPTER  III 
THE  THEORY   OF   BEAMS.     BEAM   DESIGN 

29.  Stress  and  Strain.  —  In  the  discussion  of  the  theory  of 
beams  it  is  important  to  distinguish  between  stress  and  strain; 
that  is,  it  must  be  remembered  that  the  change  of  form  which 
a  load  produces  in  a  body  is  called  the  strain,  or  deformation, 
of  the  body  due  to  the  load;  and  that  the  internal,  or  molecular, 
resistance  which   the  material  of  a  body  interposes  to  resist 
deformation  is  called  the  stress. 

30.  Coefficient  of  Elasticity  and  Hooke's  Law.  —  Within  the 
elastic  limit  of  all  materials  the  stresses  are  proportional  to  the 
strains,  but  since  the  same  intensity  of  stress  does  not  produce 
the  same  strain  in  different  materials  we  must  have  some  definite 
means  of  expressing  the  amount  of  strain  produced  in  a  body  by 
a  given  stress.     The  means  employed  is  to  assume  the  body  to 
be  perfectly  elastic  and  then  state  the  intensity  of  stress  neces- 
sary to  strain  the  body  by  an  amount  equal  to  its  own  length. 
This  stress  is  known  as  the  Modulus  of  Elasticity,  or  the  Coeffi- 
cient of  Elasticity,  and  is  denoted  by  E.     The  values  of  E  for 
different  materials  have  been  determined  and  are  practically 
the  same  for  tension  and  compression.     The  mean  values  for  E 
in  pounds  per  square  inch  for  the  materials  most  commonly 
used  in  engineering  are  as  follows:     Timber,    1,500,000;    cast 
iron,  15,000,000;  wrought  iron,  25,000,000;  steel,  30,000,000. 

There  are  no  materials  of  construction  that  are  perfectly 
elastic,  and  but  few  that  will  stretch  o.ooi  of  their  lengths  and 
remain  elastic ;  but  within  the  elastic  limit  all  bodies  are  assumed 
to  be  perfectly  elastic,  and  if  that  limit  be  not  exceeded  the 

62 


THE  THEORY   OF   BEAMS  —  BEAM   DESIGN 


coefficient  of  elasticity  is  the  ratio  of  the  unit  stress  to  the  unit 
strain.  This  is  in  accordance  with  the  discovery  made  in  1678 
by  Robert  Hooke  and  is  known  as  Hookers  law.  This  law 
expresses  the  fact  that  the  ratio  of  the  unit  stress  to  the  unit 
deformation  is  a  constant  quantity,  known  as  the  Modulus  of 
Elasticity.  Expressed  as  a  formula,  we  have 

Unit  stress 


Unit  deformation 

Cast  iron,  cement,  and  concrete  are  among  the  few  materials 
that  do  not  conform  to  Hooke's  law,  their  coefficients  of  elas- 
ticity varying  under  different  stresses;  but  it  is  customary  in  all 
investigations  to  assume  the  truth  of  the  law,  modifying  results 
by  factors  of  safety.' 

In  the  case  of  shear  stress  the  modulus  is  known  as  that  of 
transverse  elasticity  or  coefficient  of  rigidity,  commonly  denoted 
by  G.  The  mean  values  for  G  in  pounds  per  square  inch  for  the 
common  engineering  materials  are:  Timber,  150,000;  cast  iron, 
5,500,000;  wrought  iron,  10,500,000;  steel,  12,500,000. 

31.  The  Determination  of  E.  —  If  y  denotes  the  strain  pro- 
duced in  a  body  by  a  stress 


S,  and  Y  the  strain  produced 
by  the  stress  necessary  to 
stretch  the  body  to  double 
its  length,  then,  since  the 
stresses  are  proportional 
to  the  strains,  we  shall 
have  from  the  stress-strain 
diagram,  Fig.  61, 


I 


~  =  —  i     whence    E  = 

JL  Hi 


SY 


Fig.  61. 

SL 

—  i 

y 


since  the  strain,  or  deformation,  Y,  is  equal  to  L. 


64 


THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 


32.  Neutral  Axis.  —  When  a  beam,  Fig.  62,  is  subjected  to  a 
bending  moment  its  axis  is  deflected  from  its  original  straight 
position  into  one  of  curvature,  known  as  the  elastic  curve.  The 
fibers  in  the  horizontal  layers  on  the  convex  side  are  in  tension, 
and  those  on  the  concave  side  in  compression.  There  must 
therefore  be  a  layer,  or  surface,  of  fibers  that  is  neither  in  ten- 
sion nor  in  compression,  called  the  neutral  surface,  and  the  line 


Fig.  62. 

in  which  this  surface  intersects  the  plane  of  a  section  of  the 
beam  is  the  neutral  axis  of  the  section.  The  neutral  axis  inva- 
riably passes  through  the  center  of  gravity  of  a  section  whose 
material  has  its  modulus  of  elasticity  the  same  in  tension  as  in 
compression. 

Consider  the  portion  of  the  beam  included  between  the 
sections  AB  and  CD  to  be  bent  to  the  arc  of  a  circle,  and  let  r 
be  the  radius  of  the  neutral  surface,  and  c  the  distance  from  the 


THE  THEORY  OF  BEAMS  —  BEAM   DESIGN  65 

neutral  surface  to  the  outermost  surface.  The  length  of  the 
neutral  surface  will  not  change  in  the  bending,  hence  rd  denotes 
the  original  length  of  the  outermost  surface  expressed  in  radians, 
and  (r  +  c)  6  expresses  the  length  of  the  outermost  surface  after 
bending,  hence 

Deformation  =  (r  +  c)  6  —  rO  =  cQ. 
But  we  have  seen  that 

Deformation     _  S_ 
Original  length      E 
hence 

rt  =  £  =  S  (    } 

/y1  Q* 

That  is,  --  =  -  =  unit  stress  at  the  distance  unity  from  the 

Y  C 

neutral  surface,   S   being   the   unit         

stress,  tension  or  compression,  of 
the  fibers  at  the  outermost  surface, 
and  c  the  distance  from  the  neutral 
surface  to  the  outermost  surface. 

33.'  Resisting  Moment.  —  Sup- 
pose a  beam  section,  Fig.  63,  to 
be  made  up  of  a  great  number 
of  layers  a,  #1,  02,  etc.,  distant  Flg-  63> 

y>  yij  yz,  etc.  from  the  neutral  axis.     The  unit  stresses  at  distances 

S          S  S 

y,  yi,  yz,  etc.,  are  -  X  %  -  X  yi,  -  X  yz,  and  the  total  stresses 
c  c  c 

S  S  S 

on  the  elemental  areas  are  -Xay,-X  cayi,  -  X  dzyz,  etc.    The 

c  c  c 

S  S  S 

moments  of  these  stresses  are  -  X  ay2,  -  X  a\y\,  ~  X  (hy*2,  etc. 

c  c  c 

The  sum  of  all  these  moments  will  be  the  resisting  moment  of 

the  section,  hence 

S  SI 

Resisting  moment  =  -  (ay2  +  a\y^  +  Ozyz2  +[etc-X=  — > 
c  c 


Neutral  Axis 


66  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

in  which  7  is  the  moment  of  inertia  of  the  section  about  the 
neutral  axis.  The  bending  moment  must  be  equal  to  the  resist- 
ing moment,  since  for  equilibrium  the  internal  stresses  must 
be  equal  to  the  external  forces;  therefore 

SI 

Bending  moment  —  M  —  • —  (n) 

0 

SI 
The  equation  M  —  —  is  the  fundamental  equation  for  beam 

0 

investigations,  and  since  /  is  expressed  in  bi-quadratic  inches, 
M  must  be  expressed  in  pounds-inches.  The  length  of  a  re- 
quired beam  and  the  load  to  which  it  is  to  be  subjected  being 
known,  we  can  readily  find  the  maximum  bending  moment,  and 
then  by  assuming  the  allowable  working  stress,  5,  per  unit  of 

M      I 
area  at  the  outermost  fiber,  we  shall  have  —  =  -•     The  numeri- 

o          C 

cal  value  thus  obtained  for  —  must  be  satisfied  by  the  dimensions 

of  the  cross  section  of  the  beam.  The  smaller  value  of  St  whether 
for  tension  or  compression,  should  be  taken. 

34.   Section  Modulus.  —  The  factor  -   contains  the  dimen- 

c 

sions  of  the  section  and  is  a  comparative  measure  of  its  strength. 
It  is  called  the  modulus  of  the  section  and  is  denoted  by  Z. 
From  equation  (n)  we  have 

M      I  M 

-  =  -=Z,    whence    S--. 

Substituting  this  value  of  S  in  equation  (10),  we  have 

c      M  ,,      EZc      El  ,    . 

-=  —  >    whence    M  =  -   -  =  — >  (12) 

r      EZ  r          r 

which  is  the  equation  of  the  elastic  curve,  showing  the  relation- 
ship between  the  bending  moment  at  any  section  and  the  radius 
of  curvature  of  the  beam. 


THE  THEORY  OF   BEAMS  —  BEAM   DESIGN  67 

35.  Assumptions  in  the  Theory  of  Beams.  —  In  the  theory  of 
beams  there  are   three  important  assumptions  made:  (a)  That 
the  strain  increases  directly  as  the  distance  from  the  neutral 
axis;  (b)  that  the  stress  varies  as  the  strain;  (c)  that  the  coeffi- 
cient of  elasticity  is  the  same  for  tension  as  for  compression. 

No  one  of  these  assumptions  is  perfectly  true,  but  for  nearly 
all  materials  they  are  so  nearly  true  that  for  practical  purposes 
they  may  be  regarded  as  true,  so  long  as  the  elastic  limit  is  not 
exceeded. 

36.  Beam  Design.  —  Beams   are  divided  into  two  general 
classes:  Those  of  uniform  strength  and  those  of  uniform  cross 
section. 

Beams  of  uniform  strength  are  so  shaped  that  the  unit  stress 
in  the  surface  fiber  is  the  same  at  all  cross  sections.  They  are 
employed  when  it  is  desired  to  reduce  to  a  minimum  the  weight 
of  the  material  in  a  beam.  In  their  design  the  section  modulus 
must  vary  directly  with  the  bending  moment,  since  the  unit 
fiber  stress  remains  constant. 

Beams  of  uniform  cross  section  are  those  ordinarily  used,  as 
they  are  readily  sawed  from  wood  or  rolled  from  steel.  In  such 
beams  the  section  modulus  remains  constant  and  the  fiber  stress 
varies  directly  with  the  bending  moment. 

A  beam  is  designated  by  its  depth,  and  its  design  is  a 
question  of  determining  the  sectional  dimensions  which  will 
bear  with  safety  a  given  load,  the  length  of  the  beam  being 
known. 

If  the  dimensions  of  the  beam  are  to  be  determined  from  the 
maximum  allowable  fiber  stress,  the  bending  moment  at  the 
dangerous  section  must  be  computed,  after  which  the  section 

I  SI 

modulus,  -  ,  may  be  found  from  the  equation  M  = If  it  is 

c  c 

desired  to  use  one  of  the  standard  types  manufactured  by  the 
different  steel  companies,  the  beam  in  the  manufacturers'  tables 


68  THE   ELEMENTS  OF  MECHANICS  OF  MATERIALS 

having  an  -  equal  to  or  next  greater  than  the  one  found  is  to  be 
c 

selected. 

The  standard  I  beams  so  extensively  used  in  structural  work 
are  rolled  in  light,  intermediate,  and  heavy  weights  of  thirteen 
sizes.  The  different  manufacturers  issue  handbooks  containing 
tables  of  the  properties  of  the  beams  they  produce,  and  while 
there  is  an  agreement  in  sizes  as  to  depth,  there  are  marked 
differences  in  the  proportions  of  cross  sections,  and  therefore 
also  of  weight  per  foot  and  of  moments  of  inertia. 

The  upper  and  lower  horizontal  parts  of  an  I  section  are  called 
the  flanges,  and  the  vertical  part  connecting  them  is  called  the 
web.  In  the  design  of  an  I  beam  the  web  is  assumed  to 
resist  the  vertical  shear,  the  flanges  alone  resisting  the  bending 
moment. 

For  the  support  of  loads  beyond  the  capacity  of  a  single  I  beam, 
two  or  more  of  them  may  be  bolted  together  side  by  side,  in 
which  case  cast-iron  separators  are  used,  fitting  neatly  between 
the  flanges  and  bolted  to  the  webs  so  as  to  unite  the  beams 
forming  the  girder  and  cause  them  to  act  together  in  resisting 
the  load.  These  separators  should  be  placed  at  each  end  of  the 
girder  and  at  points  where  loads  are  concentrated.  For  uniform 
loading  the  spacing  of  the  separators  should  not  be  greater  than 
twenty  times  the  width  of  the  smallest  beam  flange  in  order 
to  prevent  failure  by  buckling  at  the  upper  flanges,  which  are 
in  compression. 

For  the  support  of  still  greater  loads  a  useful  and  economical 
section  can  be  formed  by  the  combination  of  two  I  beams 
having  plates  riveted  to  their  top  and  bottom  flanges,  forming 
what  is  known  as  a  box  girder. 

Should  the  cross  section  of  a  beam  be  other  than  those  listed 
as  standard,  its  dimensions  may  be  determined  by  first  computing 
the  maximum  bending  moment  and  then,  knowing  the  safe 


THE  THEORY  OF  BEAMS  —  BEAM  DESIGN       69 

allowable  value  of  S,  one  dimension  is  assumed  and  the  other 
computed. 

EXAMPLE  I.  — A  yellow  pine  beam  of  18  feet  span  is  to  support 
concentrated  loads  of  1200  pounds  and  1500  pounds  at  points 
4  feet  and  10  feet  from  the  left  support.  Taking  the  ultimate 
tensile  strength  of  yellow  pine  as  9000  pounds  per  square  inch, 
determine  the  cross-section  dimensions  of  the  beam. 


1 

10'  /-TN                           -,, 

CD 

"  CD 

u^d 

<                    in'                   •> 

A 

.  Ra 

Fig.  e>4.- 

Solution.  —  From  Fig.  64  we  have 

18  RI  =  1200  X  14  +  1500  X  8,    whence    RI  =  1600  pounds; 
also 

1 8  R2  =  1500  X  10  +  1200  X  4,    whence    R2  =  noo  pounds. 

The  construction  of  the  shear  diagram  shows  the  shear  to  pass 
through  zero  at  the  section  under  the  i5oo-pound  load,  thus 
locating  the  dangerous  section.  Then 

^/max  =  1600  X  10  —  1200  X  6  =  8800  lbs.-ft.  =  io5,6oolbs.-ins. 
Using  a  factor  of  safety  of  10,  we  shall  have 

0   9000  .  . 

o  =  •* =  900  pounds  per  square  inch 

10 

as  the  fiber  stress  at  the  outermost  surface.     Assuming  a  breadth 

of  5  inches,  and  remembering  that  c  =  -  >  we  have 

2 

M      105,600      I       bd*       $d2  , 

~  =  —     —  =  -  = =  >*—  >    whence     a  =  1 2  nearly, 

o  900          c      12  c        6 

so  that  a  5-inch  by  1 2-inch  beam  would  appear  safe. 

We  shall,  however,  investigate  the  influence  of  the  weight  of 
the  beam  itself. 


70  THE   ELEMENTS   OF  MECHANICS   OF  MATERIALS 

Yellow  pine  weighs  40  pounds  per  cubic  foot,  hence 
Weight  of  beam  =  5  X  I2  *  **£  I2  X4°  =300  pounds 

=  1  6.  66  pounds  per  foot. 

Then  RI  becomes  1600  +  150  =  1750  pounds,  and  we  shall 
have 

Mmax  =  1750  X  10  -  1200  X  6  -  16.66  X  10  X  5  =  9466.66  Ibs.-ft. 
=  113,600  pounds-inches. 

r™  0      Me      113,600  X  6  X  12  „ 

Then        S  =  —  =  —    --  -  --  =  947  ibs.  per  square  inch, 
I  5  X  1720 

which  is  greater  than  the  assumed  safe  stress  of  900  pounds,  so 
a  5.  5-inch  by  i2-inch  beam  will  be  tried.     Then 

\ir   -    i  A.        £     i.  5-5  X  12   X  l8  X  12   X  40 

Weight   of   beam  =  ^  -  —  -  =  330   pounds 

=  18.33  pounds  per  foot,  and  RI  =  1765  "pounds.     Then 

Mmax=  1765  X  10  -  1200  X  6  -  18.33  X  10  X  5=  9533.5  Ibs.-ft 
=  114,402  pounds-inches.     Then 

0      Me      114,402  X  6  X  12      0,  .     . 

5  =  —  =  —  —  -  -  -  =  867  pounds  per  square  inch, 
1  5-5  X  1720 

a  result  which  shows  the  5.5-inch  by  1  2-inch  beam  to  be  safe. 

EXAMPLE  II.  —  What  Cambria  I  beam  should  be  selected  for 
a  floor  bearing  a  load  of  180  pounds  per  square  foot,  the  beams 
to  have  a  span  of  25  feet,  to  be  spaced  8  feet  apart,  and  to  have 
a  maximum  unit  stress  of  16,000  pounds  per  square  inch? 

Solution.  - 

Load  on  each  beam  =  25  X  8  X  180  =  36,000  pounds  =  3  '°°O 

25 
=  1440  pounds  per  foot  run.     RI  =  18,000  pounds,  and 

^max  =   l8,000  X  12.5  X  12  —  1440  X  12.5  X  6.25  X  12 

=  1,350,000  pounds-inches. 


Then  =       -  =  84.37. 


THE   THEORY   OF   BEAMS  — BEAM   DESIGN  71 

A  reference  to  the  Cambria  handbook  shows  that  the  1 5-inch 
special  I  beam  of  65  pounds  per  foot  of  the  Cambria  Steel  Com- 
pany should  be  selected,  its  tabulated  modulus  being  84.8. 

EXAMPLE  III.  —  What  is  the  proper  size  of  I  beam  to  carry  a 
load  of  35,000  pounds  concentrated  at  the  middle  of  a  span  of 
25  feet,  the  fiber  stress  not  to  exceed  16,000  pounds  per  square 
inch? 

Solution.  — 

Jkfmax  =  17,500  X  12.5  X  12  =  2,625,000  pounds-inches. 

I      M      2, 62^,000 

-  =  —  =         :»     -  =  I64. 

c       S         16,000 

This  value  of  the  section  modulus  would  indicate  that  a  24-inch 
standard  Cambria  I  beam  of  80  pounds  per  foot  should  be  se- 
lected, its  tabulated  section  modulus  being  173.9.  Its  tabulated 
moment  of  inertia  is  2087.2. 

We  shall  investigate  the  effect  of  the  weight  of  the  beam  itself. 

Weight  of  beam  =  80  X  25  =  2000  pounds;  therefore  R\ 
=  18,500  pounds. 

Then  Mmax  =  18,500  X  12.5  X  12  -  80  X  12.5  X  6.25  X  12 
=  2,700,000  pounds-inches. 

0      Me      2,700,000  X  12 
5  =  T  2087.2  =  I 

which  is  500  pounds  less  per  square  inch  than  the  safe  allowable 
stress  at  the  outermost  fiber.  The  24-inch  standard  Cambria  I 
beam  weighing  80  pounds  per  foot,  is,  therefore,  the  proper 
selection.  (The  safe  loads  for  beams  given  in  the  handbooks 
issued  by  the  different  steel  companies  include  the  weight  of  the 
beam.) 

EXAMPLE  IV.  —  Heavy  1 2-inch  Cambria  I  beams  of  20  feet 
span  are  to  be  used  to  support  a  floor  bearing  a  uniform  load, 
including  the  weight  of  the  beam,  of  200  pounds  per  square  foot. 
The  outermost  fiber  stress  is  not  to  exceed  16,000  pounds  per 
square  inch.  Find  the  proper  spacing  of  the  beams. 


72  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

Solution.  —  From  the  Cambria  handbook  we  find  for  the  beam 
selected, 

Section  modulus  =  -  =41. 
c 

Let  x  denote  the  distance  between  two  consecutive  beams. 
Then 

Load  on  one  beam  =  20  x  X  200  =  4000  x  pounds,  and 

T  ,  4000  x 

Load  per  foot  =  -       -  =  200  x  pounds, 

RI  =  2000  x  pounds. 
-^max  =  2000  #  X  io  X  1  2  —  2oo  #XioX5Xi2  =  i  2o,ooo  x  Ibs.-ins. 


™,  I        M  1  20,000  X  , 

I  hen      -  =  —)    or    41  =  —  -  -  ,    whence   x  =  15.47  feet. 
c       S  16,000 

PROBLEMS 

1.  A  rectangular  beam  9  inches  deep,  3  inches  wide,  supports  a  load 
of  0.5  ton  concentrated  at  the  middle  of  an  8-foot  span.     Find  the  maxi- 
mum fiber  stress.  Ans.   0.3  ton  per  sq.  in. 

2.  A  square  timber  beam  of  1  2-inch  side  weighs  50  pounds  per  cubic 
foot,  is  20  feet  long,  and  supports  a  load  of  2  long  tons  at  the  middle  of  its 
span.     Calculate  the  fiber  stress  at  the  middle  section. 

Ans.    1037  Ibs.  per  sq.  in. 

3.  Light   i2-inch  Cambria  I  beams  of  2o-feet  span  are  to  support  a 
floor  bearing  a  uniform  load  of  175  pounds  per  square  foot.     The  outer- 
most fiber  stress  is  not  to  exceed  16,000  pounds  per  square  inch.     At  what 
distance  apart  should  the  beams  be  placed  ?  Ans.   5.5  ft. 

4.  What  safe  uniform  load  may  be  placed  on  a  wooden  cantilever  5 
feet  long,  3  inches  wide,  and  5  inches  deep  in  order  that  the  fiber  stress 
shall  be  900  pounds  per  square  inch?  Ans.   71  Ibs.  per  ft. 

5.  What  safe  uniformly  distributed  load  may  be  placed  on  a  standard 
Cambria  channel   beam  weighing  20  pounds  per  foot,  the  span  being  16 
feet,  the  web  placed  vertical,  and  the  maximum  fiber  stress  not  to  exceed 
12,500  pounds  per  square  inch?  Ans.    11,150  Ibs. 

6.  Find  the  factor  of  safety  of  a  heavy  1  2-inch  Cambria  I  beam  of  15 
feet  clear  span  when  sustaining  a  uniformly  distributed  load  of  30  tons. 

Ans.   2. 

7.  Compare  the  strength  of  a  steel  beam  of  rectangular  section,  5  inches 
by  2  inches,  with  that  of  a  9-inch  I  beam  of  the  same  material,  length,  and 
area  of  section,  the  moment  of  the  inertia  of  the  I  beam  being  111.8. 

Ans.    I  beam  is  three  times  as  strong. 


CHAPTER  IV 
THE  DEFLECTION   OF  BEAMS 

37.  Deflection  of  Beams.  —  In  Chapter  III  the  question  of 
beam  design  was  considered  from  the  standpoint  of  the  maximum 
allowable  fiber  stress,  but  the  safe  load  for  a  beam  to  carry  may 
equally  well  be  determined  from  its  maximum  allowable  deflec- 
tion. For  the  protection  of  plastered  ceilings  in  buildings,  the 
floor  beams  are  limited  in  deflection  to  about  ^^  of  their  length, 
and  their  design,  therefore,  is  based  on  considerations  of  deflection 
rather  than  of  strength. 

The  radius  of  curvature  of  any  plane  curve  is  given  in  the 
differential  calculus  as 


r  = 


dx2 


In  the  case  of  a  beam  the  curvature  is  assumed  to  be  small,  and 
therefore  the  slope  of  the  tangent  at  any  point  is  small.     In 

- 


other  words,  -    being  very  small,    -J  is  infinitesimal  and  may 
dx 

be  neglected,  and  therefore 


dx2 

Substituting  this  value  of  r  in  equation  (12)  of  Art.  34,  p.  66, 
we  have 


74 


THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 


which  is  the  equation  of  the  elastic  curve  for  the  investiga- 
tion of  the  deflection  of  beams.  In  any  particular  case  the 
value  of  M  must  be  expressed  in  terms  of  x,  and  then,  after 
two  integrations,  the  deflection,  y,  will  be  known  for  any  value 
of  x. 

38.  Points  of  Inflection.  —  Simple  beams  that  are  free  at 
their  end  supports  —  that  is,  having  no  restraining  influence 
to  keep  their  supported  ends  horizontal  —  bend  concave  up- 
ward, but  beams  that  are  fixed  at  their  end  supports  —  that 
is,  having  their  ends  built  in  so  that  the  built-in  part  is 
constrained  to  remain  horizontal  —  are  subjected  to  a  com- 
bined curvature  of  bending  which  is  concave  downward  near 
the  supports  and  concave  upward  toward  the  middle.  A  beam 
that  overhangs  its  supports,  such  as  that  of  the  Example,  p.  51, 


w 

=  200 

<        A.'      > 

W=390 

1-'                   1-        6'          , 

C 

i) 

;  6 

A 

k~~  ~~--4-      -~K' 

I- 

:  —  "  —  "L 

Fig.  65. 

Chap.  II,  reproduced  in  Fig.  65,  is  also  subjected  to  a  combined 
curvature  of  bending  between  the  supports.  The  overhangs 
exert  a  constraining  influence  to  keep  horizontal  the  parts  of 
the  beam  resting  on  the  supports  and  thus  compel  a  bending 
of  the  beam  concave  downward  near  to  and  inside  each  support. 
This  is  clearly  indicated  by  the  dotted  curves  of  Fig.  65.  In 
these  cases  of  double  curvature  in  bending,  the  points  at  which 
the  contrary  curves  separate  and  at  which  there  is  no  bending 
whatever,  are  known  as  points  of  inflection  or  points  of  contrary 
flexure.  They  can  be  found  by  placing  the  general  expression 


THE   DEFLECTION   OF   BEAMS  75 

for  the  bending  moment  within  their  limits  equal  to  zero.  Thus, 
in  Fig.  65,  it  is  evident  from  inspection  that  there  will  be  a  point 
of  inflection  between  the  left  support  and  Wz,  and  one  between 
the  right  support  and  W^.  The  expression  for  the  bending 
moment  at  any  section  within  the  limits  of  the  left  support  and 
Wz  is 

M  =  RiX-Wi(x  +  4)--(x  +  8)2. 

2 

Placing  this  expression  equal  to  zero,  and  substituting  730  for 
RI,  200  for  Wi,  and  20  for  w  (see  p.  51),  the  resulting  quadratic 
gives  a  value  of  4.42  for  x,  which  shows  there  is  a  point  of  in- 
flection at  4.42  feet  to  the  right  of  the  left  support.  The  ex- 
pression for  the  bending  moment  at  any  section  within  the 
limits  of  W2  and  the  right  support  is 

M  =  RlXl  -  Wi  (x,  +  4)  -  W2  (xl  -  12)  -  -  (Xl  +  8)2. 

2 

Placing  this  expression  equal  to  zero  and  substituting  390  for 
W2  and  500  for  R%,  the  resulting  quadratic  gives  for  Xi  the  value 
of  17.027,  showing  the  other  point  of  inflection  to  be  at  the  sec- 
tion distant  about  i  foot  to  the  left  of  the  right  support.  These 
points  of  inflection  are  indicated  at  /  and  u  in  the  bending-moment 
diagram  of  Fig.  53,  p.  52,  at  which  points  there  are  no  bending 
moments. 

The  determination  of  the  points  of  inflection  is  very  important 
in  a  mechanical  sense  as  furnishing  the  logical  location  for 
expansion  joints  in  bridge  construction. 

39.  Examples  in  Beam  Deflection.  —  The  expressions  for  the 
maximum  deflections  of  beams  under  several  different  condi- 
tions of  support  and  loading  will  here  be  derived;  those  for  other 
conditions  may  be  found  by  similar  processes. 

Example  I.  Simple  beam  with  a  load  W  concentrated  at  the 
middle,  Fig.  66. 


76 


THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 


For  any  point  between  the  left  support  and  the  middle  of  the 
beam,  we  have 


Then 
Integrating  once, 


dv       Wrr2 
El  -f  =  —  +  C. 

ax         4 


Fig.  66. 

To  find  C  we  must  know  the  slope  at  some  point.  At  the 
middle  of  the  beam  the  tangent  to  the  elastic  curve  is  hori- 
zontal, consequently  there  is  no  slope  at  that  point,  hence 

dy  L  WL2 

-7^  =  0     when    x  =  — ,   therefore  C  = • 

ax  2  16 

^  vTdy      Wx2      WL2 

Then  El-f  =- —  • 

ax         4          16 

Integrating  the  second  time,  we  get 

WL2x 


Ely  = 


Wx* 


C'. 


12  16 

At  the  supports  there  is  no  deflection,  therefore 

y  =  o    when    x  =  o  /.  C'  —  o. 
WL2x 


Then 


EIy  = 


12 


16 


The  deflection  is  a  maximum  when  #  =  —  ,  therefore 

2 


Then 


96          32 


48 


48 


THE   DEFLECTION  OF   BEAMS 


77 


Example  II.  —  Cantilever  with  a  load  W  concentrated  at  the 
end,  Fig.  67. 


Fig.  67. 

For  any  section  of  the  beam  we  have 

M  =  -W(L-x), 

Then  EI^=-W(L-x} 


There  is  no  slope  at  the  wall,  therefore  -~  =  o  when  x 

dx 

:.     C  =  o. 

Then  EI—r-  = 


=  o, 


6  2 

There  is  no  deflection  at  the  wall,  therefore  y  —  o  when  x  =  o, 


Then 


EIy  = 


6  2 

The  deflection  is  a  maximum  when  x  =  L,  therefore 


3  El 


THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 


Example  III.  —  Simple  beam  uniformly  loaded  with  w  pounds 
per  unit  of  length,  Fig.  68.     For  any  section  of  the  beam  we  have 

,,      wLx      wx2 
M  = . 

2  2 

wLx      wx2 


Then 


El 


dx2  2  2 

.  dy  _  wLx2      wx*       ~ 
dx~~      ~(> 


W  Ibs.  per  unit  of  length 


•U7L 


Fig.  68. 

The  tangent  to  the  elastic  curve  is  horizontal  at  the  middle  of 

the  beam,  therefore  -p  =  o  when  x  =  —  •     For  x  =  —  >  we  have 
dx  22 

~  _  wL3      wL3  _      wL3 
48         16  24 

„.  -n-rdy      w  Lx2      wx3      wL3 

Then 


El^f  =  - 

dx          4 


Then 


o          24 
__  wx*  _  wL3x       ~f 

12  24  24 

y  =  o  when  x  =  o,  consequently  C'  =  o. 

w 

y  = 


(2Lx*  -x4  -  Dx). 


24  El 
The  deflection  is  a  maximum  at  the  middle  of  the  beam,  where 


Then   ;ymax  =  — -    . 

24£/\4       ID 

which  W  =  wLj  the  whole  load. 


384  El  ~       384 


in 


THE   DEFLECTION   OF   BEAMS 


79 


Example  IV.  —  Cantilever  uniformly  loaded  with  w  pounds 
per  unit  of  length,  Fig.  69. 


Fig.  69. 
For  any  section  of  the  beam 

M  =  -  -  (L  -  x)2. 


Then 


o  when  x  =  o,  consequently  C  =  o. 


Then 


Then 


dx2          2 

dx          2 

dy_ 

dx 

dx  = 


2  \     2  3  I2/ 

y  =  o  when  x  =  o,  consequently  C'  =  o. 
w 

y  =  - 


24  El" 
The  deflection  is  a  maximum  when  x  =  L,  therefore 

wL*          WL*  ?  .n  whkh  w  =  wL  is  the  whole  Joad^ 


8  El 


8£/ 


Example  V.  —  Beam  built-in  or  fixed  at  the  ends  and  uni- 
formly loaded,  Fig.  70. 

This  is  a  condition  not  heretofore  encountered.  The  beam 
is  fixed  at  the  ends  in  the  sense  that  the  parts  of  the  beam  built 


8o 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


in  are  constrained  to  remain  horizontal  while  the  beam  is  being 
bent,  but  it  is  understood  that  the  beam  is  free  endwise.  The 
reaction  of  the  wall  in  keeping  the  built-in  part  horizontal  intro- 
duces the  couple,  P,  P. 


W  Ibs.  per  unit  of  length 


Fig.  70. 

The  bending  moment  at  any  section  between  the  left  support 
and  the  middle,  and  distant  x  from  the  left  support,  is 


M  =  R&  +  Px  -  P  (x  +  z)  - 


wx2 


—  +  Px  -  Px  -  Pz  -  — 

2  2 

W  (  T  2\  r> 

{J^X  —  X  )  —  Jr  Z. 

2 


Then  EIdtf=2 

dx         2  \    2 

From  the  nature  of  the  support  there  is  no  deflection  at  the  wall, 

hence  -*  =  o  when  x  =  o,  consequently  C  =  o.     To  find  the 
ax 

value  of  Pz,  which  is  the  bending  moment  at  the  support,  we 

must  know  the  value  of  -*•  at  some  other  section.     At  the 

ax 

middle  of  the  beam  the  tangent  to  the  elastic  curve  is  horizontal, 


THE   DEFLECTION  OF   BEAMS  8l 

therefore  -^  =  o  when  x  =  -  •     Substituting  this  value  of  x  in 
dx  2 

the  first  derivative,  we  get 


w(D  D\ 
—  (  ---  1 
2\8  247 


PzL  wL2 

=  o.  whence  Pz  =  - 
2  12 


the  bending  moment  at  the  support. 

Substituting  the  value  of  Pz,  and  integrating  the  second  time, 
we  have 


2\    6  I2/  24 

At  the  support  y  =  o  and  x  =  o,  therefore  C'  =  o;  and  y  is  a 
maximum  when  #  =  -  >  consequently 

2 

w    ( L^       Z/4       7A  wZ,4  W L? 

yma*  =  2^E7  (48      192      96;  =    ~  384^7  =   ~  384^7' 

in  which  W  =  wL  is  the  whole  load. 

Substituting  the  value  of  Pz  in  the  expression  for  M,  and  letting 

x  =  — '  we  have 
2 


,,      w/L2     L2\      wL2      wL2 
M  =  -( — ) = ' 

2\2  4/  12  24 


showing  the  bending  moment  at  the  middle  to  be  but  half  that 

at  the  support,  hence 

,,  wL2      WL 

Mma*=:^7:  77' 

For  the  points  of  inflection  we  place  the  second  derivative 
equal  to  zero,  thus: 


wfr         ON      w  L2       , 

-  (Lx  -  x2)  --  =o,  or  Lx  -  x2  =  —  >  whence  x  =  - 

2  12  6  6 

40.  Continuous  Beams.  —  The  beams  heretofore  considered 
have  been  either  cantilevers  or  beams  having  but  two  supports. 
A  beam  resting  on  more  than  two  supports  is  termed  a  continu- 


82 


THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 


ous  beam.  The  chief  difficulties  in  the  treatment  of  continu- 
ous beams  are  the  determination  of  the  support  reactions  and  the 
bending  moments  at  the  supports.  When  these  are  known  the 
bending  moment,  shear,  and  deflection  at  any  section  may  be 
determined. 

When  considering  beams  having  two  supports  the  support 
reactions  were  easily  found  by  means  of  two  equilibrium  equa- 
tions. With  parallel  forces  there  can  be  but  two  equilibrium 
equations,  so  that  if  a  beam  has  more  than  two  supports  some 
other  means  must  be  adopted  for  finding  the  support  reactions. 
The  whole*  treatment  of  continuous  beams  is  simplified  by  means 
of  Clapeyron's  theorem  of  three  moments,  and  is  fully  set  forth 
in  any  complete  treatise  on  the  mechanics  of  engineering.  Only 
the  special  case  of  a  beam  uniformly  loaded  and  resting  on  three 
supports  equally  spaced  and  on  the  same  level  will  here  be 
considered. 

41.  Beam  Resting  on  Three  Supports.  —  The  beam  of  Fig.  71 
is  uniformly  loaded  with  w  pounds  per  unit  of  length  and  rests 
on  three  supports  equally  spaced  and  on  the  same  level. 


w  Ibs.  per  unit  of  length 


Fig.  71. 


In  finding  the  support  reactions  we  will  first  assume  the 
middle  support  to  be  removed.  In  such  case  we  have  found 
(Art.  39,  Ex.  Ill)  that  the  deflection  at  the  middle  due  to  the 

r  WJJ 

whole  load  wL  is  -~ — —  • 
384  El 

Let  Wf  denote  the  load  on  the  middle  support.    Then  the 


THE  DEFLECTION  OF   BEAMS  83 

upward  deflection  caused  by  the  reaction  at  the  middle  support 

W'L3 

due  to  the  concentrated  load  W'  is,  by  Art.  39,  Ex.  I,———- 

48  hi 

The  tops  of  the  supports  being  on  the  same  level,  the  upward 
deflection  must  equal  the  downward  deflection,  so  that 

W'D       $wL*  .J7,      $wL 

=  -3—    -  ,     whence     W  = 

48  El      384  £/  8 

That  is,  the  middle  support  bears  |  of  the  whole  load;  and  since 
the  load  is  uniform,  each  of  the  end  supports  bears  T3s  of  the  load. 

Thus  R,  =  R3=  $™L,        and     R2  =  ^- 

16  8 

The  bending  moment  at  any  section  between  the  left  and 
middle  supports,  and  distant  x  from  the  left  support,  is 


d3) 

Then  E/g  - 

•njdy      1^1*,       w*-      n 

rLL  — -  =  — — (-  u. 

dx         2  6 

There  is  no  slope  at  the  middle  support,  therefore 

dy  L 

-r-  =  o    when    x  =  -  > 
a#  2 

consequently 

T^L 


6          2       "   48         128  384 

np-i  T'T^y  K-lX  1UX?  WLi 

i  nen  jbi  -f-  — > 

dx        2          6        384 


wx*      wUx 
— 
24        384 

the  constant  of  integration  being  zero,  because  y  =  o  when 


nT        R&*      wx*      wUx  ,    . 

Ely  =  —±  -----  —  '  (14) 

6         24        384 


x  =  o. 


84  THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 

From  equation  (14)  the  deflection,  y,  may  be  determined  for 
any  section.  Thus,  at  the  section  midway  between  the  left  and 

middle  supports,  x  =  —  and  we  have,  since  RI  =  ^        ? 
4  16 

wL*       wL4       wL*  WL* 

Ely  =  —  -----  =  —  _  > 
2048      6144      1536          3072 

WD 

3072  El, 

in  which  W  =  wL  is  the  whole  load.  This  value  of  y  is  not  the 
maximum  deflection,  as  that  does  not  take  place  at  the  middle 
of  the  span. 

To  find  the  point  of  inflection  between  the  left  and  middle 
supports  we  place  the  expression  for  the  bending  moment  within 
those  limits  [equation  (13)]  equal  to  zero,  thus 

*  wLx      wx*  ,  3  L 

---  =  o,     whence    x  =  ^—-  • 
16  2  8 

Hence  there  is  a  point  of  inflection  distant  three-eighths  the 
length  of  the  beam  from  the  left  support,  or  one-eighth  of  the 
length  of  the  beam  to  the  left  of  the  middle  support;  there  is 
another  point  of  inflection  similarly  situated  to  the  right  of  the 
middle  support. 

To  find  the  bending  moment  at  the  middle  support,  let  x  =  — 

in  equation  (13)  and  we  have 

WL 


The  greatest  bending  moment  between  the  supports  will  be 
at  the  section  where  the  shear  passes  through  zero.  We  locate 
this  section  by  placing  the  first  ^-derivative  equal  to  zero;  thus, 
by  differentiating  (13), 

dM 

—  =  RI  -  wx, 
dx 

$—-  --  wx  =  o,     whence  x  =  *-r  ; 
16  16 


THE   DEFLECTION  OF   BEAMS  85 

that  is,  the  maximum  bending  moment  between  the  supports 

from  the  left  support.     To  find 


o  jr 
occurs  at  the  section  distant  * — 


16 


its  value  we  substitute  ^-— for  x  in  (13),  and 


-*ff       u  -w±^r       u  'WJLT       g  wu~       O  WL 

M  = —  2—  -  = =  2 • 

256         512         512          512 

This  result  is  numerically  less  than  that  found  for  the  bending 
moment  at  the  middle  support;  hence  the  maximum  bending 
moment  is  at  the  middle  support,  and  we  shall  have  for  safe 

SI      WL 

loading,  —  = •    The  bending  moment  and  shear  diagrams 

c         32 

are  readily  drawn,  as  shown  in  Fig.  72. 

•C 


Fig.  72. 

42.  The  Strength  and  Stiffness  of  Beams.  —  The  direct  meas- 
ure of  the  strength  of  a  beam  is  the  load  it  will  safely  support, 
and  varies  inversely  as  the  outermost  fiber  stress  at  the  dangerous 

section.     Since  at  the  dangerous  section  S  =  — ,  in  which  M 


86  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

is  the  maximum  bending  moment,  it  follows  that  -  -  is  the  ex- 

Mc 

pression  for  the  comparative  strength  of  beams.  The  expression 
for  the  maximum  bending  moments  obtained  in  preceding  pages 

W  L 

is  M  =  -  —  ,in  which  m  has  the  values  i,  2,  4,  8, and  12,  depending 

on  the  conditions  of  support  and  loading.     The  expression  for 

the  strength  of  a  beam  may  then  be  written  —    -  ;  that  is,  the 

WLc 

strength  of  a  beam  varies  directly  as  the  section  modulus  and 
inversely  as  the  length.  In  the  special  case  of  beams  of  rec- 
tangular section,  where  7  =  —  and  c  =  -  ,  we  have  — -  as  the 

12  2  6 

value  of  the  section  modulus  -  ,  so  that  the  strength  of  such  beams 

c 

varies  directly  as  the  breadth  and  directly  as  the  square  of  the 
depth.  Doubling  the  breadth  of  a  beam  of  rectangular  section, 
therefore,  doubles  its  strength;  doubling  the  depth  increases  its 
strength  four  times.  This  furnishes  the  reason  for  placing 
rectangular  beams  with  their  greatest  dimension  vertical. 

The  ratio  of  the  maximum  deflection  of  a  beam  to  its  length  is 
termed  the  stiffness  of  the  beam,  and  its  measure  is  the  load  the 
beam  can  carry  with  a  given  deflection.  The  expression  for 

WU 
the  maximum  deflections  obtained  in  this  chapter  is  y  = 


in  which  n  has  the  values  3,  8,  48,  *- ,  and  384,  depending  on 
the  conditions  of  support  and  loading.  From  this  expression 
we  get  W  =  — j~  y  nence  the  stiffness  of  a  beam  varies  directly 

1  ^ 

as  E  and  7,  and  inversely  as  the  cube  of  the  length.  The  follow- 
ing table  furnishes  a  comparison  of  beams  under  different  con- 
ditions of  loading  and  support,  the  relative  strength  and  stiffness 
of  each  being  expressed  in  comparison  with  that  of  the  weakest 


THE    DEFLECTION  OF   BEAMS 


beam,  which  is  the  cantilever  with  a  concentrated  load  at  the 
free  end: 


Kind  of  beam  and  nature  of  load  and 
support. 

Maximum 
bending 
moment. 

Maximum 
deflection. 

Relative 
strength. 

Relative 

stiffness. 

Cantilever,  load  at  free  end  

WL 

wu 

I 

I 

Cantilever,  load  uniformly  distrib- 
uted   

WL 

2 

3  El 

WL3 

8  El 

2 

a| 

WL 

WU 

Simple  beam,  load  at  middle  

Simple  beam,  load  uniformly  dis- 
tributed   

Beam  with  fixed  ends,  uniformly 

4 

WL 

8 

WL 

48  El 

$WL3 
384  El 
WL3 

4 

8 

16 

2Sl 

loaded  

7QA  77y 

12 

120 

PROBLEMS 

i.   Find  the  expression  for  the  maximum  deflection  of  a  beam  fixed  at 
the  ends  and  with  a  load  W  concentrated  at  the  middle.     Find  the  points 


of  inflection. 


WLZ       L 

Ans.  -       —  ;  -  from  each  end. 
192  hi     4 


2.  Find  the  maximum  deflection  of  a  24-inch  Cambria  I  beam,  25  feet 
long  and  weighing  80  pounds  per  foot,  when  resting  on  end  supports  and 
bearing  a  load  of  35,000  pounds  at  the  middle.     Neglect  the  weight  of  the 
beam.  Ans.   0.314  in. 

3.  A  rolled  steel   beam,  symmetrical   about   its   neutral  axis,   has  a 
moment  of  inertia  of  72  inch  units.     The  beam,  which  is  8  inches  deep,  is 
laid  across  an  opening  of  10  feet  and  carries  an  evenly  distributed  load  of 
9  tons,  including  its  own  weight.     Find  the  maximum  deflection;  find  also 
the  maximum  fiber  stress. 

Ans.   0.218  inch;  7.5  tons  per  s,q.  in. 

4.  A  lo-inch  Cambria  I  beam,  the  moment  of  inertia  of  which  is  tabu- 
lated as  1 2 2. i,  is  laid  across  an  opening  of  16  feet.     In  addition  to  a  concen- 
trated load  of  1000  pounds  at  the  middle  it  carries  a  uniformly  distributed 
load  of  14,400  pounds,  including  the  weight  of  the  beam.     Find  the  maxi- 
mum deflection,  taking  E  as  29,000,000;  find  also  the  maximum  fiber 
stress.  Ans.   0.416  in.;  16,120  Ibs.  per  sq.  in. 

5.  A  beam  of  wood,  8  inches  wide  and  12  inches  deep,  and  fixed  at 
the  ends,  covers  a  span  of  14  feet.    It  bears  a  uniformly  distributed  load 


88  THE  ELEMENTS  OF  MECHANICS   OF   MATERIALS 

of  10,000  pounds,  including  its  own  weight.    Find  the  maximum  deflec- 
tion, taking  E  as  1,500,000;  find  also  the  maximum  fiber  stress. 

Am.   0.072  in.;  729  Ibs.  per  sq.  in. 

6.  Find  the  safe  uniformly  distributed  load  for  a  6-inch  I  beam  rest- 
ing on  end  supports  20  feet  apart  if  the  deflection  is  limited  to  siff  of 
span.    E  =  30,000,000;  I  =  24;  and  weight  of  beam  is  14.75  pounds  per 
linear  foot.  Ans.    118. 6  Ibs.  per  linear  foot. 

7.  How  much  stronger  is  a  beam  4  inches  wide,  8  inches  deep,  and  8 
feet  long,  weighing  71  pounds,  than  one  3  inches  wide,  5  inches  deep,  and 
14  feet  long,  weighing  58  pounds?  Ans.  4.9  times. 


CHAPTER  V 
COLUMNS.     SHAFTS 

43.  Columns.  —  A  column,  or  strut,  is  a  straight  beam 
acted  on  compressively  at  its  extremities,  and  of  such  length 
compared  with  its  diameter,  or  least  sectional  dimension,  that 
failure  will  result  from  buckling  or  lateral  bending,  instead 
of  by  crushing  or  by  splitting.  In  addition  to  the  many 
familiar  applications  of  columns  in  structural  work,  the  pis- 
ton rods  and  connecting  rods  of  steam  engines  are  classed  as 
columns. 

If  columns  were  initially  absolutely  straight,  made  of  homo- 
geneous material,  and  exactly  centrally  loaded,  there  would  be 
no  difference  in  the  character  of  their  failure  from  that  of  short 
specimens.  These  three  conditions  are  never  fulfilled  in  practice, 
and  in  consequence  columns  are  weaker  than  short  blocks  of  the 
same  material. 

No  satisfactory  theoretical  discussion  of  columns  has  been 
made,  and  all  the  formulae  used  in  their  design  contain  con- 
stants determined  by  experiment.  The  formula  having  the  most 
rational  basis  is  the  one  attributed  to  Rankine  or  to  Gordon, 

which  is 

AS 


in  which  W  is  the  load  on  the  column  expressed  in  pounds,  A  the 
sectional  area  of  the  column  in  square  inches,  S  the  ultimate 
compressive  strength  of  a  short  block  of  the  material,  L  the 
length  of  the  column  in  inches,  K  the  least  radius  of  gyration  of 

89 


THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 


the  section,  and  a  a  constant  quantity  determined  by  experiment 
for  different  materials. 

The  strength  of  a  column  depends  largely  upon  the  manner 
in  which  its  ends  are  secured.  If  the  ends  are  flat  the  column 
is  said  to  have  square  or  flat  ends;  if  one  end  be  fixed  and  the 
other  end  hinged,  as  in  the  case  of  a  piston  rod,  the  column  is 
said  to  be  fixed  at  one  end  and  free  at  the  other;  if  both  ends  are 
hinged,  as  in  the  case  of  a  connecting  rod,  the  column  is  said  to 
be  free  or  round  at  the  ends. 

The  value  of  K~  in  Rankine's  formula  is  found  from  the  rela- 
tion /  =  AK2.  The  values  for  S  in  pounds  per  square  inch, 
for  a,  and  for  suitable  factors  of  safety  for  the  three  conditions 
of  bearing,  are  given  in  the  following  table : 


Timber. 

Cast  iron. 

Wrought 
iron. 

Structural 
steel. 

Hard  steel. 

5 

8000 

84,000 

55-ooo 

6o,000 

120,000 

a 
(square  ends) 

3OOO 

sj.OOO 

36,000 

36,000 

25  ooo 

a 
(fixed  and  free)  

1690 

2,810 

20,250 

24,000 

14,060 

a 
(free  ends)  

75° 

1,250 

9,000 

9,OOO 

6,2^0 

Factor  of  safety 
(buildings).            .        .  . 

8 

8 

6 

4 

cr 

Factor  of  safety 
(bridges) 

10 

10 

e 

7 

Factor  of  safety 
(shocks) 

ic 

i  c 

IO 

jr 

It  will  be  observed  from  the  table  that  the  ratios  of  the  values 
of  a  for  the  square  ends  to  the  values  for  fixed  and  free  and  for 
free  ends  are  as  1.75  to  i  and  as  4  to  i  respectively,  and  these 
ratios  indicate  the  relative  strengths  of  long  columns  with  those 
end  conditions. 


COLUMNS  QI 

Example  I.  —  A  hollow  cylindrical  cast  iron  column  16  feet 
long,  with  square  ends,  sustains  a  load  of  200,000  pounds  when 
used  in  a  building.  Outside  diameter,  10  inches;  inside  diameter, 
8  inches.  Is  it  safe? 

Solution. — 

A  =  0.7854  (100  —  64)  =  28.27  square  inches. 

K2=  L  =  *  (I0>000  -  4096)  =  io 

A  64  X  28.27 

Substituting  in  Rankine's  formula,  we  have 

s  =  go^ooo  r          d6XI2)M 
28.27    \_        5000  X  io.24J 

pounds  per  square  inch.     The  factor  of  safety,  then,  is  - 
=  6.9,  showing  the  column  to  be  unsafe. 

Example  II. —  Find  the  safe  load 'for  a  1 2-inch  Cambria  I 
column  weighing  31.5  pounds  per  foot,  16  feet  long,  and  with 
square  ends,  the  column  to  be  used  in  a  building. 

Solution.  —  Referring  to  the  Cambria  handbook,  we  find  that 
A  =  9.26  square  inches,  and^  =  i.oi.  The  factor  of  safety  is  4. 

„,  T,_  9.26  X  15,000 

Then  W  =  — 2 — -— -^ — ^ —  =  69>3°°  pounds. 

,  ylO  X  12)" 

36,000  X  (i.oi)2 

44.  Design  of  Columns.  —  In  designing  a  column  we  have 
given  the  form  of  its  section,  its  length,  the  material  of  which  it 
is  to  be  made,  the  load  it  is  to  carry,  and  the  manner  of  securing 
its  ends.  The  problem  is  then  to  determine  the  necessary  area 
of  cross  section.  The  general  method  of  procedure  is  to  deter- 
mine from  the  data  given  the  necessary  cross-section  area  for  a 
short  block  of  the  material;  then,  knowing  that  the  section  area 
of  the  required  column  must  be  larger,  assume  dimensions  which 
give-  a  larger  area,  and  then,  by  means  of  Rankine's  formula, 
ascertain  the  unit  stress  resulting  from  such  assumption.  If  it 
is  too  great,  the  assumed  area  is  too  small;  if  too  little,  a  smaller 


92  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

area  must  be  chosen.    Proceed  thus  by  trial  and  error  until  a 
satisfactory  solution  is  obtained. 

Example  III.  —  A  column  of  timber  of  square  section,  15  feet 
long  and  with  square  ends,  is  required  to  support  a  steady  load 
of  10  tons.  Find  the  necessary  dimensions  of  the  section. 


c 

Solution.  —  Using  a  factor  of  safety  of  8,  we  have  -  =  1000 

8 

pounds  per  square  inch  as  the  safe  working  unit  stress. 

2000  X  10 

Then  -  =  20 

1000 

square  inches  area  of  section  needed  for  a  very  short  column,  or 
one  less  in  length  than  ten  times  the  least  dimension  of  the 
section.  As  the  required  section  must  be  larger,  we  will  assume 
an  area  of  36  square  inches. 

Then  *,      '      _®1_     3. 

A      12  X36 

Therefore 

0      20,000  f     .    (15  X  i2)2l  . 

S  =  —  —  —   i  +  ^-^  -  —    =  2555  pounds  per  square  inch. 
36    L         3000  X  3  J 

Since  this  is  much  larger   than   the   allowable  unit   stress  of 
1000  pounds,  we  must  assume  a  larger  area. 
Try  an  area  of  64  square  inches.     Then 


S  =  i  +  =  945  Pounds  per  square  inch. 

04    L         3°°°  X  5  J 

This  result  is  a  trifle  small,  so  we  will  try  a  square  section 
whose  side  is  7.9  inches.     Then 

(7-9)4 


5  -     1  +       =  986  pounds  per  square 


COLUMNS  93 

which  is  quite  near  the  allowable  unit  stress;  therefore  a  column 
of  square  section,  having  a  side  of  7.9  inches,  will  suffice. 

Sometimes  all  the  dimensions  may  be  assumed  except  one, 
and  then,  after  expressing  A  and  K2  in  terms  of  the  unknown 
dimension,  we  can  substitute  them  in  Rankine's  formula  and 
solve  the  resulting  bi-quadratic  for  the  unknown  dimension. 
Thus,  in  the  problem  just  solved,  let  x  denote  the  length  of  the 
side  of  the  unknown  square  section.  Then 

r4  r2 

A=x\     and     K*=-^-  =  —, 

12  X2         12 

and 

I000  _  ^aooor  (I5  X  12)'  X  12-1  _  20,000_  /  I^ 

x2     I  3000  a;2        J  #2       \  x2   / 

whence  x  =  7.9  inches. 

Example  IV.  —  A  hollow  cylindrical  cast  iron  column  20  feet 
in  length  is  required  to  sustain  a  steady  load  of  164,000  pounds. 
Determine  its  cross-sectional  dimensions. 

Solution.  —  Using   a   factor  of   safety  of   8,  we  have   4'000 

8 

=  10,500   pounds   per   square   inch  as  the  safe  working  unit 

stress. 

Then — ~ =  15.62  square  inches  needed  for  a  very  short 

10,500 

column. 

Assuming  an  area  of  cross  section  of  25  square  inches,  and 
assuming  further  that  the  outside  diameter  of  the  column  shall 
be  10  inches,  we  shall  have,  calling  d  the  inside  diameter, 

25  =  0.7854  X  ioo  -  0.7854^, 
whence  d  =  8.256  inches. 

Then  g-j-'K^-fr'^-Kxsi. 

A  64  X  25 

and 

0    164,000  r       (20  x  i2)2  "I 

5  =  -^-[i  +  gooo  x  IQ  SJ  =  '3,75°  Pounds  per  sq.  in., 


94  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

which  is  greater  than  the  allowable  unit  stress  of  10,500  pounds. 
The  cross-sectional  area  must  therefore  be  larger,  so  we  will 
assume  an  area  of  33.75  square  inches.  Then 

33-75  =  0.7854  X  ioo  -  0.7854  d2, 
whence  d  =  7.55  inches, 

and  K2  =  9.82. 

Then 

c       1  64,000  T  (20  X  i2)2~j 

o  =  -  i  +  -  -^-~  =  10,560  pounds  per  square  inch, 

33-75   L         5000X9-82] 

a  result  sufficiently  near  the  allowable  unit  stress  to  warrant 
making  the  column  10  inches  in  outside  diameter  and  the  metal 
1.25  inches  thick. 

Example  V.  —  Using  a  factor  of  safety  of  4,  what  safe  load 
can  be  supported  by  a  7-inch  Cambria  channel-and-plate  column 
14  feet  long  and  weighing  34.8  pounds  per  foot? 

Solution.  —  From  the  Cambria  handbook  the  section  area 
of  the  column  is  found  to  be  10.2  square  inches,  and  the  least 
radius  of  gyration  2.63.  Then 


W  = 


4 iQ.2  X  12,500 

~1&~  (14  X   I2)2  =  28,224 

aK2  36,000  X  (2.63)2  36,000  X  6.9 


=  114,500  pounds. 


PROBLEMS 


1.  A   hollow   cylindrical   cast  iron   column   with   square   ends,  whose 
thickness  of  metal  is  2  inches,  length  24  feet,  and  outside  diameter  24 
inches,  is  to  be  used  in  a  building.     What  safe  load  can  it  sustain  ? 

Ans.   1,141,000  Ibs. 

2.  A  round,  solid,  cast  iron  column  with  square  ends  is  15  feet  long 
and  6  inches  in  diameter.     What  safe  load  can  it  carry  ? 

Ans.    76,500  Ibs. 

3.  A  1 2 -inch  Cambria  channel-and-plate  column  used  in  a  building  is 
18  feet  long  and  weighs  64.8  pounds  per  foot.     What  safe  load  can  it 
carry?  Ans.   223,300  Ibs. 


SHAFTS  95 

4.  Find  the  external  diameter  and  thickness  of  metal  of  a  hollow  cast 
iron  column  that  will  safely  carry  a  load  of  194,000  pounds,  its  length 
being  16  feet,  and  the  ratio  of  the  inside  and  outside  diameters  being  0.8. 
Use  a  factor  of  safety  of  8. 

Ans.    10.5  ins.;  i.i  ins. 

5.  Find  the  thickness  of  metal  of  a  hollow  column  of  structural  steel, 
6  inches  in  diameter  and  28  feet  long,  to  support  a  load  of  100,000  pounds 
in  a  building.  Ans.  0.775  m- 

45.  Shafts.  —  A  shaft  in  mechanics  is  a  revolving  bar  used 
for  the  transmission  of  power  generated  by  an  engine  or  other 
motor.     In  the  design  of  a  shaft  the  important  consideration  is 
the  nature  of  the  stress  to  which  it  is  to  be  subjected.     In  the 
form  of  shaft  commonly  known  as  an  axle,  where  the  load  is 
applied  transversely,  the  stress  is  chiefly  due  to  bending;  in 
transmission  shafting,  such  as  that  used  in  machine  shops  and 
factories,  the  stress  is  principally  due  to  torsion;  in  heavier  forms, 
such  as  crank  shafts,  the  stress  is  one  of  combined  torsion  and 
bending. 

Since  rotating  cylindrical  bodies  have  equal  strength  in  all 
positions,  shafts  are  usually  circular  in  section.  We  shall  be 
concerned  only  with  those  shafts  of  circular  section  that  are 
subjected  chiefly  to  a  torsional  stress,  and  with  those  subjected 
to  a  combined  stress  of  torsion  and  bending. 

46.  Torsion.  —  When  shafts  are  used  for  the  transmission  of 
power  they  are  subjected  to  a  shearing  stress  due  to  a  torque  or 
twisting  moment,  the  measure  of  which  is  the  product  of  the 
applied  force  and  the  distance  of  the  point  of  its  application 
from  the  axis  of  the  shaft.     If   P  denotes  the  applied  force 
and  r  the  perpendicular   distance  of  its  point  of   application 
from  the  axis  of  the  shaft,  then  Pr  is  the  measure  of  the  twist- 
ing moment. 

When  a  uniform  circular  shaft  is  subjected  to  a  twisting 
moment  each  element  parallel  to  the  axis,  such  as  AB  of  Fig.  73, 
undergoes  a  helical  deformation  AD,  and  each  elemental  area 


96 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


of  a  section  is  subjected  to  a  shearing  stress  produced  by  a  strain 
called  torsion,  the  stress  varying  directly  as  the  distance  of  the 
elemental  area  from  the  axis.  The  angle  6  through  which  a 
longitudinal  element  is  twisted  is  called  the  angle  of  shear,  and 
varies  directly  as  the  distance  of  the  element  from  the  axis. 
The  angle  a  is  called  the  angle  of  twist,  and  is  proportional  to 
the  length  of  the  shaft. 


Fig-  73- 

Suppose  a  section  of  length  dx,  Fig.  74,  to  be  cut  from  a  shaft 
by  planes  perpendicular  to  the  axis,  and  let  da  be  the  angle  of 

twist  for  this  section.  The  angle  of 
twist  varying  with  the  length  of  the 
shaft,  we  shall  have 

dx 


Fig.  74. 


da.          , 
=  —  >     whence 


adx 

T 


,  cda       Ca 

whence     6  =  — —  =  — > 
dx       L 


Expressing  6  and  da  in  radians,  we  have 
BD  =  6dx  =  cda, 

c  being  the  radius  of  the  shaft. 
From  Art.  30,  p.  62,  we  have 

Unit  stress _  5 

=  ~9 


Unit  deformation 


whence 


S  =G9  = 


Gca 


S  being  the  unit  stress  at  the  outermost  surface. 


SHAFTS 


97 


Suppose  the  section  of  the  shaft,  Fig.  75,  to  be  made  up  of  a 
great  number  of  small  areas,  a,  a\,  #2,  etc.,  distant  y,  yi,  y2,  etc., 
from  the  axis.  Each  of  these  small  areas  is  subjected  to  a  shear- 
ing stress  s,  si,  s2,  etc. 


Fig.  75- 

If  6*  denotes  the  unit  stress  of  the  material  at  the  outermost 
fiber,  distant  c  from  the  axis,  it  is  the  maximum  unit  stress  to 
which  the  material  will  be  subjected,  since  the  stresses  vary 
directly  as  the  distance  from  the  axis. 

The  resisting  moment  of  the  section  will  be  the  sum  of  the 
moments  of  the  elemental  resistances,  as,  a^i,  a^s^  etc.,  and 
since  there  must  be  equilibrium  between  the  internal  and  external 
forces,  the  resisting  moment  will  equal  the  twisting  moment  and 
we  shall  have 

Mt  =  asy  +  aiSiyi  +  OzSzyz  +  etc. 
But 


S      s      Si      $2         ,  Sy 

—  =  -  =  —  =  — '     whence    s  =  —  >  Si  = 

c     y     yi     y%  c  c 


Then 


C 

S 

=  -  (ay2 
c 


etc.)  = 


SI 


in  which  Ip  is  the  polar  moment  of  inertia. 


98  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

rpi  c       Mtc      Gca 

Inen  5  =  -y—  =  —  —  , 

IP        L 

MtL 
whence  a  =  -——  radians. 


For  a  circular  section     /„  =  ---     Hence 

32 


Angle  of  twist  in  degrees  -  ^ 

in  which  D  is  the  diameter  of  the  shaft. 

For  a  hollow  circular  shaft  of  external  diameter  D  and  internal 


7T  ~ 


~ 

diameter  0,    Ip  =  —  -  -  >     and 
32 


Angle  of  twist  for  hollow  shaft  =     $S*Mt^- 

G  (D*  —  a4) 


57 

From  the  equation  Mt  =  —  -  >  we  get 

c 


in  which  Zp  denotes  the  polar  modulus  of  the  shaft  section. 
It  is  thus  seen  that  the  strength  of  a  shaft  to  resist  torsion  varies 
as  its  polar  section  modulus,  just  as  the  strength  of  a  beam  to 
resist  bending  varies  as  its  plane  section  modulus  (see  Art.  34, 
p.  66). 

From  what  has  preceded  it  is  known  that  the  least  plane 
moment  of  inertia,  7,  of  a  circle  is  just  one-half  its  polar  moment 
of  inertia,  7P.  That  Ip  should  be  greater  than  7  might  have 
been  expected,  since  the  material  of  a  circular  shaft  is  in  far 
better  form  to  resist  torsion  than  to  resist  bending. 

For  shafts  of  small  diameter  and  comparatively  long  lengths, 
such  as  line  shafting  in  mills  and  shops,  the  angle  of  twist  may 
be  too  great  to  insure  safe  transmission.  In  such  cases  the  diam- 
eter of  the  shaft  should  be  determined  with  reference  to  the 


SHAFTS  99 

stiffness  of  the  material  rather  than  to  its  strength,  the  deter- 
mining factor  being  the  angle  of  twist. 

The  angle  of  twist  should  enter  into  the  design  of  shafts  under 
ten  inches  in  diameter,  and  it  may  be  accepted  as  a  result  of 
experience  that  if  the  angle  of  twist  does  not  exceed  i°  in  18  diam- 
eters of  length,  shafts  are  sufficiently  stiff. 

s-i 

We  have  shown  that  5  =  — - ,  in  which  a  is  expressed  in  radians, 

LJ 

and  c  —  —  -    With  the  limiting  value  of  L  =  18  D  for  i°  of 

2 

twist,  we  shall  have 


57.3  X  i8  D      2063 

Taking  Gas]  10, 500,000  for  wrought  iron  and  as  12,500,000  for 
steel,  we  have  for  light  shafting: 

For  wrought  iron,  S  =  5000  pounds  per  square  inch. 
,  For  steel,  5  =  6000  pounds  per  square  inch. 

Example.  —  A  steel  shaft  2  inches  in  diameter  and  10  feet 
long  is  subjected  to  a  twisting  moment  of  7500  pounds-inches. 
What  is  the  angle  of  twist?  Is  the  shaft  safe? 

Solution.  — 

A       i        r  A.     •  *.         584  AfjL         584X7^00X10X12  ,    0 

Angle  of  twist  =  —         -   = L2—  =  2.63  . 

GD4  12,500,000  X  1 6 

AH        ui          i      r  *.    •  4.  Length  10  X  12          0 

Allowable  angle  of  twist  = r^         -  =  — =  3-3  • 

1 8  diameters        18  X  2 

The  shaft  is  therefore  safe. 

PROBLEMS 

1.  A  3-inch  steel  shaft,  200  feet  in  length,  is  subjected  to  a  twisting 
moment  of  35,000  pounds-inches.     Is  it  safe?  Ans.   Unsafe. 

2.  If  the  shaft  of  problem  i  is  unsafe,  what  should  be  its  diameter  to 
insure  safety?  Ans.   3.07  ins. 


100  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

47.  Combined  Torsion  and  Bending.  —  In  cases  of  heavy 
shafting,  such  as  propeller  shafts  of  vessels,  the  weight  of  the 
shaft  occasions  a  bending  action  which  increases  the  stress 
in  the  material  over  that  due  to  torsion  alone.  The  effect 
of  this  combined  action  of  torsion  and  bending  is  not  exactly 
determinable  in  practice,  owing  to  the  inexact  computation  of 
the  bending  moment  involved.  It  may  be  shown,  however, 
that  if  M  and  Mt  denote  the  bending  and  twisting  moments 
respectively  at  any  section  of  a  shaft,  and  Me  the  equivalent 
twisting  moment  that  would  produce  the  same  intensity  of 
stress  as  that  due  to  the  combined  torsion  and  bending,  we 
shall  have 


Mf. 

The  mean  twisting  moment  of  shafts  of  steam  engines  is 
determined  from  the  mean  pressure  in  the  cylinder,  and  the 
greater  the  ratio  of  expansion  of  the  steam  the  greater  difference 
there  will  be  between  the  maximum  and  minimum  moments  to 
which  the  shaft  will  be  subjected;  and  since  a  shaft  must  be  strong 
enough  to  resist  the  maximum  stress  to  which  it  may  be  liable, 
the  twisting  moment  expressing  the  maximum  stress  due  to  the 
combined  twisting  and  bending  should  be  the  basis  of  the  cal- 
culation to  determine  its  diameter.  For  practical  purposes  the 
equivalent  twisting  moment  may  be  taken  as  1.5  times  the 
mean  twisting  moment,  and  in  some  exceptional  cases  this 
multiplier  may  be  increased  to  2. 

48.  Transmission  of  Power  by  Shafts.  —  If  P  denotes  the 
mean  force  acting  on  the  crank  pin  of  an  engine  at  a  distance  of 
r  inches  from  the  axis  of  the  shaft,  then  Pr  is  the  mean  twisting 
moment  of  the  shaft  in  pounds-inches. 

For  N  revolutions  of  the  shaft  per  minute  the  path  of  P  is 

2  irrNP 
2  vrN  inches,  and  the  work  performed  is  -        -  foot  pounds. 


SHAFTS  10  I 

Then 

Horse  power  transmitted  =  H.P.  = 


12X33,000 
I2  X       °°°  X  H.P. 


whence     Pr  =  Mt- 

2irN  N 

We  have  shown  that 


,,       SIP  ,  cn, 

Mt  =  —  2  =  —  —  =  0.196  SZ)3, 
c  16 

in  which  D  is  the  diameter  of  the  shaft. 

For  short  and  heavy  shafts  where  the  angle  of  twist  is  not 
important,  the  value  of  S  may  be  taken  as  10,000  pounds  per 
square  inch  for  steel  and  as  8000  for  wrought  iron.  Taking  the 
equivalent  twisting  moment  as  1.5  Mt,  we  shall  have  for  heavy 
steel  shafts: 

'763,025  X  H.P.  X  1.5  VST. 

=  V     0.196  X  10,000  N      =3'64V-F' 

and  for  heavy  wrought-iron  shafting,  where  S  =  8000, 


The  light  shafts  of  machine  and  wood-working  shops,  running 
at  from  125  to  250  revolutions  per  minute,  and  carrying  pulleys 
from  which  machines  are  driven,  are  subjected  to  bending  as  well 
as  torsion.  In  such  cases  it  is  in  accordance  with  good  practice 
to  take  the  equivalent  twisting  moment  at  1.5  times  the  mean, 
and  to  take  the  value  of  S  for  steel  as  6000  pounds  per  square 
inch,  giving  for  the  diameter  the  value 


H.P. 

N 

For  hollow  shafts  in  which  the  internal  diameter  d  is  one-half 
the  external  diameter  D,  the  common  practice  for  shafts  of 
steamships,  we  have 


if.- -- 0.184.^. 

c  32  c 


102  THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 

Therefore,  for  hollow  shafts  of  steel,  with  S  =  10,000  and  the 
equivalent  twisting  moment  1.5  times  the  mean,  we  have 


_     /63,025XH.P.Xi.5 
"V    0.184X10,000 TV 

For  hollow  iron  shafts,  5  =  8000,  and 


LP. 


N 

EXAMPLE.  —  A  steel  shaft  is  to  have  30  per  cent  of  its  section 
area  removed  in  making  an  axial  hole,  and  is  to  transmit  9000  H.P. 
at  1 20  revolutions  per  minute.  Taking  the  equivalent  twisting 
moment  as  1.5  times  the  mean  twisting  moment,  find  the  internal 
and  external  diameters  of  the  shaft;  show  that  the  shaft  is  25.55 
per  cent  lighter  than  a  solid  shaft  would  be  to  transmit  the  same 
power;  and  show  that  the  two  shafts  would  be  of  equal  strength. 

Solution.  —  Area  of  removed  section  =    '^  r —  . 

4 

and   d  =  0.5477/2. 
4 
o^-A7P* 

H.P. 

12  X  33,000 

whence 

n      ,_      9000X12X33,000X1.5  ,    .     , 

1.5  Pr  =  Mt  =  - — =  7, 090,000  pounds-inches. 

2?rX  120 
or         _ c (  r»4 

Mt  = 


c  $2c                             i6D 

Then  1786.78  D3  =  7,090,000, 

whence  D  =  15.83  inches, 

and  d  =  0.5477  D  =  8.67  inches. 

A  solid  steel  shaft  to  transmit  9000  H.P.  at  120  revolutions  per 
minute  would  have  a  diameter 


D  =  3.64V2 —  =  IS-3S  inches. 


SHAFTS  103 

Since  the  weights  of  shafts  are  proportional  to  their  sectional 
areas,  and  as  the  areas  vary  as  the  squares  of  their  diameters, 
we  find  the  hollow  shaft  to  be 


_  25.55percent 


lighter  than  the  solid  shaft. 
For  the  hollow  shaft 


=       =        =  0-1787  £>3  =  0.1787  X  (is.83)3  =  709- 

C  ij 


For  the  solid  shaft 


_  7r£4  __  7r£>3  =  3.1416  (I5-35)3  _ 
p      320    "   16  "  16 


The  moduli  of  the  sections  being  equal,  the  shafts  are  of  equal 
strength. 

PROBLEMS 

1.  Find  the  diameter  of  a  wrought-iron  shaft  to  transmit  90  H.P.  at 
130  revolutions  per  minute.     What  should  be  the  diameter  if  there  is  a 
bending  moment  equal  to  the  twisting  moment  ? 

Ans.   3.54  ins.;  4.75  ins. 

2.  Find  the  diameter  of  a  steel  shaft  for  a  steam  engine  having  an 
overhung  crank.     Diameter  of  cylinder,  18  inches;  mean  steam  pressure, 
130  pounds  per  square  inch;  stroke  of    piston,  30  inches;  overhang  of 
crank,  i.e.  middle  of   crank  pin  to   middle  of   bearing,   18  inches.     Use 
Me  =  M  +  VM2  +  Mf.  Ans.   9.56  ins. 

3.  A  4-inch  steel  shaft  30  feet  long  is  found  to  have  an  angle  of  twist 
of  6.2°  when  transmitting  power  while  making  130  revolutions  per  minute. 
Find  the  H.P.  transmitted.  Ans.   194.64. 

4.  Find  the  diameter  of  a  steel  shaft,  50  feet  long  to  transmit  30  H.P. 
at  210  revolutions  a  minute.  Ans.   3.2  ins. 

5.  Find  the  diameter  of  a  hollow  steel  shaft  to  transmit  10,000  H.P.  at 
120  revolutions  per  minute,  the  external  diameter  being  twice  the   inner, 
and  the  equivalent  twisting  moment  1.5  times  the  mean. 

Ans.   16.25  ins. 


104  THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

6.  A  steel  shaft  is  to  have  25  per  cent  of  its  sectional  area  removed  in 
making  an  axial  hole,  and  is  then  to  transmit  7200  H.P.  at  120  revolutions 
per  minute.    Taking  the  equivalent  twisting  moment  as  1.5  times  the 
mean,  find  the  internal  and  external  diameters  of  the  shaft,  and  show  that 
the  shaft  is  21.7  per  cent  lighter  than  a  solid  shaft  would  be  to  transmit  the 
same  power.  Ans.  14.56  ins,  and  7.28  ins. 

7.  The  main  shaft  of  a  machine  shop,  in  transmitting  40  H.P.  at  120 
revolutions  per  minute,  carries  the  average  number  of   pulleys.    Find  its 
diameter  if  made  of  steel.  Ans.  3  ins. 


CHAPTER  VI 

INTERNAL  WORK  DUE  TO  DEFORMATION.     SUDDENLY 
APPLIED   LOADS 

49.  Resilience.  —  The   stresses   thus   far   encountered   have 
been  those  produced  by  external  forces  gradually  applied  to 
bodies  under  static  conditions,  and  the  resulting  deformations 
necessarily  produced  internal  work  in  the  bodies.     As  the  stresses 
produced  were  within  the  elastic  limit,  the  bodies  resumed  their 
original  shapes  upon  the  removal  of  the  acting  forces,  and  in 
doing  so  the  internal  work  due  to  the  deformations  was,  in 
accordance  with  the  principle  of  the  conservation  of  energy, 
given  out  in  the  form  of  mechanical  energy.     It  is  thus  seen 
that  the  internal  work  of  deformation  is  a  form  of  potential 
energy,  called  resilience. 

50.  Effect  of  Suddenly  Applied  Loads.  —  The  sudden  appli- 
cation of  loads,  such  as  a  train  passing  rapidly  over  a  bridge, 
causes  greater  deflections  than  would  be  the  case  were  the  same 
loads  applied  gradually.     These  deflections  are  but  momentary, 
as  resilience  causes  a  vibratory  action  in  the  affected  bodies 
until  the  effect  of  the  shock  due  to  the  sudden  application  of  the 
load  disappears. 

If  a  load  W  be  gradually  applied  to  a  beam,  producing  a  stress 
S,  the  intensities  of  the  load  and  stress  will  gradually  increase 
from  o  to  W  and  from  o  to  S  respectively,  the  mean  values 

TTr  Q 

being  —  and  — .     If  y  be  the  deflection,  we  shall  have  for  the 

equality  of  the  external  and  internal  work 

W  9 

-  X  y  =  -  X  y,  whence  S  =  W. 

105 


106  THE   ELEMENTS   OF   MECHANICS   OF  MATERIALS 

If  the  same  load  be  applied  suddenly,  its  intensity  remaining 

constant  during  the  period   of  action,  the  resulting   stress  S' 

S' 

will  increase  from  o  to  5",  its  mean  value  being  — ,  and  a  deflec- 

2 

tion  y'  will  be  produced.     We  shall  then  have 

Wyf  =  -  X  /,    whence   S'  =  2  W. 

2 

In  other  words,  the  sudden  application  of  a  load  produces  a 
stress  twice  as  great  as  that  produced  by  the  same  load  applied 
gradually. 

Should  a  load  W  fall  on  a  beam  through  a  distance  h  and  pro- 
duce a  deflection  y,  the  kinetic  energy  developed  would  be 
W(h  +  y) ;  but  all  this  energy  would  not  be  converted  into  work 
of  deformation  on  account  of  the  inevitable  loss  of  kinetic  energy 
sustained  by  all  partly  elastic  bodies  during  impact. 

Example  I.  —  What  load  may  fall  through  a  distance  of 
8  inches  on  the  middle  of  a  1 2-inch  Cambria  I  beam  in  order  that 
the  maximum  fiber  stress  shall  not  exceed  20,000  pounds  per 
square  inch,  supposing  70  per  cent  of  the  kinetic  energy  of  the 
falling  load  to  be  converted  into  work  of  deformation?  The 
length  of  the  beam  is  16  feet,  and  its  tabulated  moment  of 
inertia  is  215.8. 

Solution.  —  Let  W  denote  the  gradually  applied  load  that 
would  perform,  in  producing  a  deflection  y,  the  same  work  of 
deformation  as  that  performed  by  the  falling  load  W' .  The 

W  Wy 

mean  value  of  W  is  — ,  and  its  work  of  deformation  is  — -• 

2  2 

WL 

The  bending  moment  at  the  middle  section  due  to  W  is > 

and  we  shall  have 

0      Me      WLc  TJ/     4^7 

S  =—=——,    whence    W=-—-> 
I         4  /  Lc 

The   deflection   due   to    W   we   have    found  to  be      — =-= , 

48  hi 

(Art.  39,  Ex.  I). 


INTERNAL   WORK  —  SUDDENLY   APPLIED   LOADS  107 

WD        SD          20,000(16  Xi2)2 

Hence     y  =    n  ^r  =  — —  = * —  =  0.341  inch. 

48  El      12  EC      1 2  X  30,000,000  X  6 

Then 

Work  of  deformation  by  W  =  ^  =  —  X  -^-  =  ^^ 

2          Lc        12  EC      6Ec2 

(20,ooo)2  X  215.8  X  16  X  12 

=  - =  2557.6  inch-pounds. 

6  X  30,000,000  X  36 

Work  of  deformation  by  W  is  0.7  W'  (8  +  0.341),  hence 

2557.6  =  0.7  TF  X  8.341, 

whence  W  =  438  pounds. 

PROBLEM 

i.  From  what  height  may  a  load  of  1200  pounds  fall  on  a  1 5-inch 
Cambria  I  beam  15  feet  long  and  having  a  moment  of  inertia  of  511,  in 
order  that  the  maximum  fiber  stress  shall  not  exceed  30,000  pounds  per 
square  inch,  and  supposing  70  per  cent  of  the  kinetic  energy  of  the  falling 
weight  to  be  converted  into  work  of  deformation  ? 

Ans.   9.37  ins. 

The  sudden  application  of  loads  produces  impact,  resulting  in 
stresses  greatly  in  excess  of  those  produced  by  the  same  loads 
when  applied  gradually,  and  in  the  recovery  from  the  shocks 
thus  occasioned  the  material  sustains  such  rapid  vibrations  that 
its  molecular  structure  and  elasticity  may  be  impaired.  In 
the  design  of  machines  and  structures  that  are  subjected  to 
shocks  the  effect  of  impact  is  an  important  consideration,  and  the 
determination  of  the  resilience  of  a  given  material  is  the  best 
measure  of  its  ability  to  withstand  shocks. 

51.  Modulus  of  Resilience.  —  If  a  bar  be  placed  in  a  testing 
machine  and  subjected  to  a  gradually  increasing  load  in  a  direc- 
tion to  produce  a  tensile  stress,  the  elongation  produced  will  be 
proportional  to  the  load,  provided  the  elastic  limit  of  the  material 
be  not  exceeded.  The  load,  or  external  force,  will  gradually 

W 

increase  from  o  to  W,  and  its  mean  value  will  be    —     If  y 

2 


108  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

denotes  the  elongation,  the  expression  for  the  work  done  is  — -  • 

2 

It  is  a  fundamental  assumption  that  the  total  stress  in  the  bar 
is  uniformly  distributed  throughout  its  section,  so  that  if  A 
denotes  the  area  of  a  section  of  the  bar  and  S  denotes  the  unit 
stress,  we  shall  have  W  =  AS',  and  it  has  been  shown  in  Art.  31, 

SL 

p.  63,  that  y  =  — -,  so  that  the  expression  for  the  work  in  this 
h, 

instance  may  be  written 

S2 
Internal  work  =  — -  •  AL. 

2E 

But  AL  is  the  volume  of  the  bar,  consequently  the  work  done  is 

proportional  to  the  volume  of  the  bar,  and  therefore  to  its  weight. 

The  work  done  in  stretching  the  bar  to  its  elastic  limit  is  its 

resilience,  and  if  S  be  taken  as  the  unit  stress  at  the  elastic  limit, 

S2 
the  ratio  — -  is  known  as  the  modulus  of  resilience.     The  resilience 

of  the  bar  is  then  found  by  multiplying  its  volume  by  the  modulus 
of  resilience. 

Example  II. — A  steel  bar  10  feet  long  and  2  inches  in  diameter 
is  stretched  to  the  elastic  limit;  what  is  its  resilience? 

Solution.  —  The  elastic  limit  of  steel  in  tension  is  50,000  pounds 
per  square  inch,  and  the  coefficient  of  elasticity  is  30,000,000 
pounds  per  square  inch. 

rr^i  AT  j  i        r       -r  S2  (so.ooo)2  250 

Then.     Modulus  of  resilience  =  — -  = ^—  —  =  — s2-  > 

2  £       2  X  30,000,000         6 

and  Volume  of  bar  =  3.1416X120  =  377  cubic  inches. 

Then,  Resilience  of  bar  =    $°      ^77  =  15,708  inch-pounds. 

PROBLEMS 

i.  In  gradually  elongating  a  steel  bar  i  square  inch  in  section  and  20 
feet  long,  work  to  the  amount  of  150  foot  pounds  is  expended.  Find  the 
applied  force  and  the  amount  of  elongation  produced. 

Ans.    21,213  Ibs.;  0.17  in. 


RESILIENCE   OF   BEAMS  109 

2.  A  wrought  iron  bar  4  inches  in  diameter  and  10  feet  long  has  its 
unit  tensile  stress  increased   from   10,000   to    20,000  pounds  per  square 
inch.     What  additional  potential  energy  is  stored  in  the  rod  by  the  oper- 
ation? Ans.   673  ft.  Ibs. 

3.  If  600  foot  pounds  of  work  are  expended  in  increasing  the  unit 
tensile  stress  of  a  steel  bar  4  inches  in  diameter  from  10,000  to  20,000 
pounds  per  square  inch  what  is  the  length  of  the  rod?  Ans.    7.5  ft. 

52.  Resilience  of  Beams.  —  The  work  performed  in  bending 
a  beam  to  the  maximum  deflection  within  the  elastic  limit  is 
the  resilience  of  the  beam.  Denoting  the  load  by  W,  the  stress 

in  the  beam  increases  from  o  to  W,  the  mean  value  being  — 

2 

The  work  performed,  or  the  resilience,  is  therefore  equal  to  half 
the  load  multiplied  by  the  deflection.  Several  illustrations  of 
the  methods  of  finding  the  resilience  of  beams  under  different 
conditions  of  loading  will  here  be  given. 

I.   Cantilever  with  load  at  extremity. — The  maximum  deflection 
WT*  WT3    W       W2T3 

i       .  VV  J-j      /  \      .  -[-»         TT\      J.1  *i*  •      VV  J^t        Vr  rr     X-/ 

being  — — :  (Art.  39,  Ex.  II),  the  resilience  is  -—  •  —  =  — —  - 
3  LI  3  LI    2       6  hi 

Mmax  =  WL  =  — ,  whence  W  =  —  • 
c  Lc 

S2I2U  S2IL          S2     K2 

Then,  Resilience  =  =  — =  — ;  •  AL, 

6  EIL2c2      2  E .  3  c2      2  E   $c2 

in  which  AK2  is  substituted  for  /. 

d  I        bd*        d2 

For  rectangular  sections,  c  =  -  and  K2  =  —  =  — —  =  — •  ,  so 

2  A       12  bd      12 

K2      i 
that  -"•  =  -.     Then,  for  cantilevers  of  rectangular  section  and 

c2       3 
with  concentrated  load  at  the  free  end,  we  have 

S2     AL 

Resilience  =  — , 

2  E     9    ' 

or  it  is  the  product  of  the  modulus  of  resilience  and  one-ninth 
the  volume  of  the  beam. 


110  THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 

II.   Simple    beam    with   load   concentrated   at   middle.  —  The 

WD 
maximum  deflection  being  -—  —  (Art.  39,  Ex.  I),  the  resilience 

40  JtLl 

.     WU    W       W2D 

S 


48  El    2       96  El 


WL      SI      .          T,,      4  SI 
=  --  =  —  •  ,  whence  W  =  -  — 
4          c  '  Lc 


i652/2L3 
Then,     Resilience  = 


96  EIL?c2      2  £  -  3  c2      2  E   3  c 

j£2      i 

For  rectangular  sections,  —  =  -  ,  and  we  shall  have  for  a  simple 

c        3 
beam  with  load  at  middle 


v    ... 

Resilience  =  —  -          , 
2  h     9 

which  is  the  same  as  that  found  for  a  cantilever  with  load  at  its 
extremity. 

III.  Cantilever  uniformly  loaded.  —  If  a  cantilever  be  loaded 
with  w  pounds  per  linear  unit,  the  elemental  load  is  wdx,  and 
if  y  be  the  corresponding  deflection  the  elemental  external  work 

is  —  —  ,  the  summation  of  which  will  be  the  resilience. 

2 

In  Art.  39,  Ex.  IV,  we  found 


24  £/ 

»L 


Then,      Resilience  -  -^--  f  (6  L2x2  -  4  Lx*  + 
48  £7  Jo 


48E/ 


[2    7-2^3  7.4     ,  _ 

"~ 


The  bending  moment  at  any  section  distant  x  from  the  wall 

w(L  -  x)2  _ 

2 

when  x  =  o,  hence 


w  (L  -  x)2          w  (L2  -  2  Lx 
is -  = -  ,   and   is   a  maximum 

2  2 


wL2      57      ,  2  SI 

Mmax  = =  — ,  whence  w  = — - 

2  C  L?C 


INTERNAL   WORK  —  SUDDENLY    APPLIED   LOADS          III 
S2      I    K2 

Then,  Resilience  =  — - .-.—-.  AL. 

2E   5    c2 

For  rectangular  sections,  —  =  -  ,  hence 

c         3 

^    .r  S2    AL 

Resilience  =  — -  •  — 
2E     15 

for  cantilevers  of  rectangular  section  and  uniform  load. 

IV.   Simple  beam  uniformly  loaded.  —  The  elemental  load  is 
wdXj  and  if  y  be  the  corresponding  deflection,   the  elemental 

external  work  is  — — ,  the  summation  of  which  will  be  the 

2 

resilience. 

In  Art.  39,  Ex.  Ill,  we  found 

y  =  — —  (2  Lx3  —  x*  —  L3x),  therefore 
24  rLl 

w2      CL 

Resilience  =  — —  /   (2  Lx3  —  x4  —  L3x)  dx 
48  El  t/o 

(70|2  |~~   /    /y»4  /y»5  7"  3/v»2    1  HOr1  T  " 

OU  J^Ji*  Ji/  JU  Ji     I  (JU  J^t 


2lL  = 
J0 


48  El  _  2        5         2     0          240  El 

The  bending  moment  at  any  section  distant  x  from  the  left 

,  .  wLx      wx2      w  ,  T  <>\        j   • 

support  is—  -  =  —  (Lx  —  x2),  and  is  a  maximum  at  the 

222 

middle  of  the  beam,  where  x  =  -  ,  hence 

2 

,,  wL2      SI      .  SSI 

Mmax  =  ——  =  —  ,  whence  w  =  -—  • 

o  C  Li  C 

Then 

T>        "I-  S*          8        K*        AT 

Resilience  =  —  -  ----  -  •  AL. 

2  £    15    c2 

For  rectangular  sections,  —  7  =  -  ,  hence 

S2     SAL 

Resilience  =  —  -  — 
2  E     45 

for  beams  of  rectangular  section  and  uniformly  loaded. 


112 


THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 


53.   Resilience  of  Torsion.  —  Suppose  the  shaft  of  Fig.  76  to 
be  subjected  to  a  twisting  moment  P'r',  and  in  consequence 

suppose  y  to  be  the 
displacement  of  an  ele- 
mental area  a  distant 
r  from  the  axis.  If  R 
denotes  the  unit  shear- 
ings tress  in  area  a, 
then 


Fig.  76. 

-ITT    i         r          i  •     T    i     •  o  +  Ra  Ray 

Work  performed  in  displacing  a  =  -         -  X  y  =  — - 

From  Art.  31  we  shall  have 


Then, 


G     R       ,  RL 

-.  =  — ,    whence  y  =  -—  • 
L      y  G 

Work  performed  in  displacing  a  = 


R2aL 

2G 


rp  .  !         ,         f          i 
Total  work  performed  = 


, 

H 


, 

| 


, 
+  etc., 


2  2  2 

which  is  an  expression  for  the  resilience  of  torsion  of  the  beam. 

If  S  denotes  the  unit  shearing  stress  in  the  remotest   ele- 

g 
mental  area,  distant  c  from  the  axis,  then  —  is  the  unit  shearing 

0 

Ql 

stress  at  a  unit's  distance  from  the  axis,  and  —  is  the  unit  shear- 

c 

ing   stress   in    the  elemental  area  a  distant  r  from  the  axis. 
Hence  R  =  —  >  and  in  like  manner  RI  =  —  >  and  R2  =  —  • 


We  shall  then  have 

Resilience  of  torsion  = 


S2L 


2GS 


2  Gc 


etc. 


2GC2 


(ar2  +  atfi2  +  a2r22  +  etc.) 


2Gc2 


SL    LAK*  -  S*     ^L     AT 
2Gm     c2      ~2G*c2          ' 


since^  the  polar  moment  of  inertia  Ip  =  AK2. 


RESILIENCE   OF  TORSION  1 13 


-,  .          ,  r  ,       ,,2  P  .     7T.D2          £>2 

For  circular  sections,  c  =  —    and    K2  =  -~  = r  =  —  > 

2  ^        32         4         8 

so  that  -7  =  -  • 

c2       2 

Then,  for  shafts  of  circular  section  we  have 

S2    AL 
Resilience  of  torsion  =  —  •  —  > 

2(r        2 

or,  it  is  the  product  of  the  coefficient  of  resilience  for  torsion  and 
one-half  the  volume  of  the  shaft. 

Comparing  the  expressions  derived  for  the  resilience  of  a  bar 
in  tension,  of  beams  of  rectangular  sections  under  different  con- 
ditions of  loading,  and  of  a  circular  shaft  under  torsion,  we  ob- 
serve that  the  resilience  of  the  bar  is  9  times  as  great  as  that  of 
a  cantilever  loaded  at  its  free  end,  9  times  that  of  a  simple  beam 
loaded  at  the  middle,  15  times  that  of  a  cantilever  uniformly 
loaded,  5!  times  that  of  a  simple  beam  uniformly  loaded,  and 
twice  that  of  a  shaft  under  torsion. 

PROBLEMS 

1.  Derive  the  expression  for  the  resilience  of   a  beam  uniformly  loaded 
with  w  pounds  per  linear  unit  and  fixed  at  the  ends.  .         S2     AL 

2E'  15 

2.  Derive  the  expression  for  the  resilience  of   a  beam  fixed  at  the  ends 
and  having  a  load  W  concentrated  at  the  middle.  ,         S*    AL  < 

2£     9 


CHAPTER  VII 
GRAPHIC    STATICS 

SYSTEM  OF  LETTERING.    FORCE  DIAGRAM.    FUNICULAR 
POLYGON 

54.  Graphic  Statics.  —  The  methods  by  which  static  problems 
are  solved  by  means  of  scale  drawings  constitute  graphic  statics. 

In  very  many  cases  the  determination  by  calculation  of  the 
forces  transmitted  through  the  different  parts  of  a  structure 
involve  tedious  and  difficult  processes,  with  the  consequent 
liability  to  error,  and  in  the  end  the  effort  to  check  the  accuracy 
of  the  results  occasions  a  procedure  as  prolonged  and  tedious 
as  the  processes  themselves.  By  the  graphic  method,  however, 
solutions  are  readily  obtained,  and  the  process  itself  furnishes  a 
check  as  to  accuracy. 

55.  System  of  Lettering.  —  To  the  system  of  lettering  dia- 
grams devised  by  A.  H.  Bow  is  due  much  of  the  facility  in 
making  graphic  solutions. 

The  two  important  features  of  the  Bow  system  of  lettering 
are:  (i)  The  placing  of  a  letter  in  each  of  the  spaces  between 
the  lines  of  action  of  the  external  forces;  (2)  the  naming  in  clock- 
wise order  of  each  force  by  the  two  letters  flanking  it. 

The  forces  of  Fig.  77  will  serve  to  illustrate  the  Bow  system. 
The  five  forces  are  in  equilibrium,  and  the  letters  A,  B,  C,  D, 
and  E  are  placed  in  the  spaces  between  them.  Other  letters, 
or  even  numbers,  would  serve  the  purpose,  and  they  might  be 
placed  in  any  order,  but  it  will  be  found  convenient  to  commence 
at  the  left  and  letter  the  spaces  alphabetically.  The  force  of 
20  pounds  is  flanked  by  the  letters  A  and  B  and  is  known  as  the 

114 


GRAPHIC   STATICS 


force  AB,  not  as  BA,  since  the  letters  must  be  read  clockwise. 
In  like  manner  the  remaining  forces  are  known  as  BC,  CD,  DE, 
and  EA. 


201bs. 


40  Ibs.  10  Ibs. 


15  Ibs. 


5  Ibs. 


Fig.  77. 


6 

Fig.  78. 


Should  the  system  of  forces  act  on  an  open-framed  structure, 
such  as  a  roof  truss,  Fig.  79,  a  letter  must  be  placed  within  each 
open  space  of  the  frame  in  addition  to  those  placed  in  the  spaces 
between  the  external  forces. 
The  external  forces,  by  the  Bow 
notation,  are  known  as  AB,  BC, 
CD,  DE,  and  EA.  The  stress 
in  the  member  connecting  joints 
i  and  2  is  known  as  BG.  The 
stress  in  the  member  separating 
the  spaces  lettered  F  and  G,  may 


Fig.  79. 


be  known  as  GF  or  as  FG,  according  to  the  end  of  the  member 
under  consideration.  If  the  end  considered  is  that  at  joint  i 
then  the  stress  in  the  member  is  known  as  GF,  because  the  forces 
about  that  joint,  taken  in  clockwise  order,  are  AB,  BG,  GF,  and 
FA.  If  the  other  end  of  the  member  is  under  consideration  the 
stress  is  known  as  FG,  because  the  forces  about  the  joint  at  that 
end,  in  clockwise  order,  are  known  as  FG,  GH,  HI,  IE,  and  EF. 


Il6          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

56.  Force  Diagram.  —  Taking  the  forces  of  Fig.  77  in  clock- 
wise order,  and  denoting  them  by  the  corresponding  small  let- 
ters of  the  alphabet,  we  may  represent  them  in  magnitude  and 
direction  in  a  diagram  by  lines  drawn  to  some  chosen  scale. 
Such  a  diagram  is  known  as  a  force  diagram.  Thus,  A  being  the 
first  letter  in  clockwise  order  from  the  left,  the  letter  a  will  be 
the  starting  point  of  the  force  diagram  of  Fig.  78,  and  since  the 
force  AB  acts  downward,  ab,  one  inch  in  length,  will  represent 
it  to  the  scale  of  20  pounds  to  the  inch.  The  force  BC,  next  in 
clockwise  order  and  equal  to  40  pounds,  acts  upward  and  will  be 
denoted  by  be,  2  inches  in  length  and  measured  upward.  In  like 
manner,  cd  measured  downward  and  one-half  inch  in  length,  de- 
notes the  force  CD  of  10  pounds;  de  measured  upward  and  one- 
quarter  inch  in  length,  denotes  the  force  DE  of  5  pounds;  and, 
finally,  ea,  which  is  found  to  measure  three-quarters  of  an  inch, 
properly  denotes  the  force  EA  of  15  pounds.  It  will  be  noted  that 
the  force  EA  is  of  just  sufficient  magnitude  to  fill,  when  drawn  to 
scale,  the  space  between  e  and  a  and  thus  close  the  force  diagram. 
This  force  diagram  is,  in  reality,  a  closed  polygon  which  has 
resolved  itself  into  a  straight  line  in  consequence  of  all  the  forces 
being  parallel.  Had  it  not  closed,  the  system  being  in  equilib- 
rium, there  would  have  been  an  error  in  the  construction. 

With  a  correct  construction,  and  the  last  point  of  a  force 
diagram  not  falling  on  the  first  point,  the  construction  would 
indicate  a  resultant  force,  equal  in  magnitude  to  the  scale  dis- 
tance between  the  last  point  and  the  first  point,  and  acting 
upward  or  downward  according  as  the  last  point  had  fallen  above 
or  below  the  first  point. 

If  one  of  the  forces  of  Fig.  77  were  unknown  it  could  be  found 
by  means  of  the  force  diagram.  Suppose  the  force  BC  unknown. 
Then,  commencing  with  the  force  CD,  the  force  diagram  would 
be  constructed  by  representing  in  clockwise  order  the  forces  CD, 
DE,  EA,  and  AB  by  the  scale  distances  cd,  de,  ea,  and  ab  respec- 


FUNICULAR   POLYGON  117 

tively,  of  Fig.  78,  c  being  the  first  point  of  the  diagram  and  b 
the  last  thus  determined.  The  missing  force  must  then  be  repre- 
sented by  be,  measured  upward,  and  as  it  measures  2  inches,  it 
correctly  does  so  for  the  force  BC  of  40  pounds. 

The  system  of  lettering  enables  the  resultant  of  any  number 
of  the  forces  to  be  read  at  once  from  the  force  diagram.  Sup- 
pose the  resultant  of  the  forces  BC,  CD,  and  DE  were  required. 
The  first  and  last  letter  in  the  naming  of  these  forces  are  B  and 
E  respectively,  so  that  be  on  the  force  diagram  represents  the 
required  resultant  in  magnitude  and  direction.  The  measure- 
ment of  be  is  found  to  be  1.75  inches,  which,  to  the  scale,  repre- 
sents 35  pounds,  the  resultant  of  the  forces  40  +  5  —  10,  and  it 
acts  upward,  since  be  is  measured  upward. 

57.  Funicular  Polygon.  —  Suppose  a  jointed  frame,  Fig.  80, 
to  be  acted  on  at  its  hinged  joints  by  a  system  of  forces  in  equi- 
librium, the  members,  bars,  or  links  of  the 
frame  being  free  to  adjust  themselves  to 
the  best  position  for  withstanding  the 
action  of  the  forces.  Such  a  figure  is 
called  a  funicular  polygon,  the  word  funic- 
ular having  no  mechanical  significance. 

Since  the  system  is  in  equilibrium,  the 
funicular  polygon  must,  of  course,  be  a 

Fig.  80. 

closed  polygon.     The  equilibrium  is  occa- 
sioned by  the  balance  between  the  internal  forces,  or  stresses, 
in  the  members  and  the  external  forces,  and  the  whole  being 
in  equilibrium,  each  joint  in  itself  is  in  equilibrium. 

Since  the  equilibrium  at  each  joint  is  the  result  of  the  action 
of  three  concurring  forces,  viz.,  the  external  force  at  the  joint 
and  the  stresses  set  up  in  the  two  members,  it  follows  that  a 
triangle  may  be  constructed  for  each  joint  which  will  represent 
these  forces  in  magnitude  and  direction. 

If,  for  example,  the  force  AB  be  known,  the  equilibrium  of  the 


Il8          THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 

joint  at  which  it  is  applied  is  maintained  by  the  action  of  the 
external  force  AB  and  the  stress  forces  in  the  members  BO  and 
OA.  Draw  ab,  Fig.  81,  parallel  to  the  force  AB,  and  make  its 
length,  to  some  chosen  scale,  represent  the 
magnitude  of  AB.  From  b  and  a  draw  lines 
parallel  to  BO  and  OA  respectively,  inter- 
secting at  o.  Then  abo  is  the  triangle  of 
forces  for  the  joint  ABO,  and  bo  and  oa 
represent  not  only  the  directions  of  the 
stresses  in  the  members  BO  and  OA,  but 
their  magnitudes  as  well,  to  the  same  scale  as  ab  represents 
the  force  AB.  The  stress  bo  in  the  end  of  the  member 
BO  at  which  AB  is  applied  occasions  an  equal  and  opposite 
stress  ob  at  the  other  end  of  BO,  and,  in  fact,  such  is  the 
case  in  all  of  the  members  of  the  polygon,  for  in  no  other 
way  could  the  equilibrium  be  maintained.  Taking  the  next 
joint  in  clockwise  order,  we  have  the  external  force  BC  and 
the  stress  forces  CO  and  OB  in  equilibrium.  But  ob  has 
just  been  found  to  represent  in  magnitude  and  direction  the 
stress  OB.  Hence,  by  drawing  from  b  and  o  lines  parallel  re- 
spectively to  BC  and  CO,  intersecting  at  c,  we  shall  have  bco 
as  the  triangle  of  forces  for  the  joint  BCO,  and  be  and  co  will 
represent  in  magnitude  and  direction  the  force  BC  and  the  stress 
in  the  member  CO  respectively.  We  now  know  the  stress  oc  at 
the  joint  CDO,  and  can  construct  the  triangle  of  forces,  cdo, 
for  that  joint.  In  like  manner  we  can  proceed  and  determine 
all  the  external  forces  and  stresses  in  the  members,  the  last  line, 
fa,  closing  the  diagram,  thus  proving  that  the  system  is  in 
equilibrium. 

It  will  be  observed  that  the  sides  of  the  polygon  just  con- 
structed represent  the  external  forces  of  the  funicular  polygon, 
and  therefore  abcdef  is  the  force  diagram.  Furthermore,  all  the 
lines  representing  the  stresses  in  the  members  of  the  funicular 


FUNICULAR   POLYGON  119 

polygon  meet  at  a  point  called  the  pole.  These  stress  lines  are 
known  as  vectors. 

From  what  has  preceded  it  is  seen  that  if  all  the  external 
forces  that  are  applied  to  a  funicular  polygon  are  known  in 
magnitude  and  direction,  and  also  the  directions  of  two  of  its 
members,  the  force  polygon  can  be  drawn.  For  the  force 
diagram  can  be  drawn  from  the  known  forces,  and  the  inter- 
section of  the  vectors  parallel  to  the  known  directions  of  the  two 
members  gives  the  pole.  The  directions  of  the  remaining  mem- 
bers are  then  found  by  drawing  the  rest  of  the  vectors. 

A  force  diagram  of  any  system  of  forces  in  equilibrium  can 
be  drawn,  and  by  choosing  any  pole,  a  funicular  polygon  with 
respect  to  that  pole  can  then  be  constructed  to  which  the  forces 
may  be  applied. 

If  the  magnitudes  and  directions  of  a  system  of  forces  acting 
on  a  body  are  known  and  the  system  is  not  in  equilibrium,  the  line 
of  action  of  the  force  required  for  equilibrium  may  be  determined 
by  means  of  the  funicular  polygon.  For  the  force  diagram  of 
the  given  forces  may  be  drawn  and  the  gap  representing  its  lack 
of  closure  will,  by  the  polygon  of  forces,  give  the  magnitude  and 
direction  of  the  resultant.  A  pole  for  this  force  diagram  may 
be  selected  arbitrarily  and  vectors  drawn.  Starting  at  a  selected 
point  in  the  line  of  action  of  any  one  of  the  forces,  a  funicular 
polygon  may  be  drawn  with  respect  to  the  chosen  pole,  and  the 
intersection  of  the  two  members  of  this  funicular  that  are  parallel 
to  the  vectors  drawn  to  the  extremities  of  the  resultant  in  the 
force  diagram  gives  the  joint  of  the  funicular  at  which  the  re- 
quired force,  equal  and  opposite  to  the  resultant,  must  be  applied. 

PROBLEMS 

i.   The   parallel    forces  of   Fig.   a    .6  ,12  ,2         .3 

are  in  equilibrium.     Find  by  a  force    I 
diagram  the  magnitude  of  the  force 


BC.  Ans.   17.  Fi& 


120          THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 


15  Ibs. 


25  Ibs. 


2.  Forces  of  20  pounds,  25  pounds,  15  pounds, 
and  30  pounds  act  on  a  body  in  the  directions 
shown  in  Fig.  b.  Find  the  magnitude  and  direc- 
tion of  the  resultant.  Find  also  by  means  of 
a  funicular  polygon  the  line  of  action  of  the 
force  required  for  equilibrium. 

Ans.   Resultant  =  n  Ibs. 


Fis-  b-  58.   Illustrations.  —  To  illustrate  prop- 

erties of  the  funicular  polygon  the  closed  polygon  of  Fig.  82 
has  been  drawn,  having  the  known  forces  AB,  BC,  CD,  DE,  and 
EA  acting  at  its  joints. 


Fig.  82. 


Fig.  83. 


Draw  to  a  selected  scale  the  force  diagram  abode  of  Fig.  83. 
The  triangle  of  forces  abo  for  the  joint  ABO  determines  the 
pole  o,  and  the  other  vectors  may  then  be  drawn. 

Let  a  plane  xy  divide  the  polygon  into  two  parts.  The  forces 
of  the  part  to  the  right  of  xy  are  AB  and  BC,  and,  by  the  triangle 
of  forces,  their  resultant  is  ac,  Fig.  83,  acting  from  a  to  c.  The 
forces  of  the  part  to  the  left  of  xy  are  CD,  DE,  and  EA ,  and,  by 
the  polygon  of  forces,  their  resultant  is  ca,  acting  from  c  to  a. 
Hence,  the  resultant  of  the  forces  of  one  part  has  the  same  mag- 
nitude and  line  of  action  as  the  resultant  of  the  forces  of  the 


FUNICULAR   POLYGON  121 

other  part,  but  acting  in  opposite  directions,  showing  that  the 
resultant  of  the  forces  of  one  part  maintains  equilibrium  with 
the  forces  of  the  other  part. 

To  find  where  the  resultant  of  the  two  forces  to  the  right  of  xy 
acts,  we  replace  the  forces  ab  and  be  in  the  force  diagram  by  their 
resultant  ac,  so  that  our  force  diagram  now  becomes  acde.  The 
vectors  oc,  od,  oe,  and  oa  of  the  force  diagram  have  o  as  their  pole, 
so  that  a  funicular  polygon  may  be  drawn  with  respect  to  o, 
having  its  sides  parallel  to  these  vectors.  We  already  have  OC, 
OD,  OE,  and  OA  of  Fig.  82  parallel  to  these  vectors,  but  they  do 
not  close  the  polygon,  and  since  a  funicular  polygon  must  close 
we  produce  OC  and  OA  until  they  intersect,  and  at  the  joint  thus 
formed  the  resultant  R,  having  the  magnitude  and  direction 
of  ac,  will  act  as  shown  by  the  dotted  lines  of  Fig.  82.  By  a 
similar  process  the  resultant,  ca,  of  the  forces  CD,  DE,  and  EA 
may  be  shown  to  act  at  the  same  joint  but  in  the  opposite 
direction. 

An  inspection  of  the  funicular  polygon  of  Fig.  82  and  of  the 
force  diagram  of  Fig.  83  shows: 

(a)  The  resultant  of  the  forces  CD,  DE,  and  EA  to  the  left 
of  the  section  xy  is  given  by  ca,  the  first  and  last  letters  of  the 
forces  when  named  in  clockwise  order;  in  like  manner  the  re- 
sultant of  the  forces  AB  and  BC  to  the  right  of  xy  is  ac. 

(b)  The  letters  in  the  force  diagram  which  name  the  resultant 
also  name  the  members  of  the  funicular  which  have  to  be  pro- 
duced to  their  intersection  in  order  to  get  a  point  in  the  line  of 
action  of  the  resultant.     Thus,  the  resultant  of  the  forces  CD 
DE,  and  EA  is  ca,  and  by  producing  the  members  C  and  A  to 
their  intersection  in  Fig.  82  a  point  in  the  line  of  action  of  the 
resultant  is  obtained. 

It  will  be  observed  that  the  introduction  of  the  external  force 
R,  the  resultant  of  the  forces  AB  and  BC,  changes  the  original 
funicular  polygon  to  one  having  fewer  joints  by  one.  Should  the 


122          THE   ELEMENTS   OF  MECHANICS   OF  MATERIALS 

members  C  and  E  be  produced  to  their  point  of  intersection 
and  the  resultant  ce  of  the  forces  CD  and  DE  be  applied 
at  the  point,  the  funicular  would  be  reduced  to  one  of  three 
joints. 

The  members  cut  by  the  section  xy  are  A  and  C,  and  the 
stresses  in  these  members,  in  magnitude  and  direction,  are  oa 
and  oc  of  Fig.  83.  On  the  right  side  of  xy  the  stresses  in  the 
members  A  and  C  act  in  the  directions  shown  by  the  arrowheads, 
and  their  resultant,  in  magnitude  and  direction,  is  ca,  which  is 
opposed  by  the  equal  and  opposite  external  force  R,  or  ac.  On 
the  left  side  of  xy  the  stresses  in  the  members  A  and  C  act  in  the 
directions  indicated  by  the  arrowheads,  and  their  resultant  is 
ac  in  magnitude  and  direction.  The  external  forces  to  the  left 
of  xy  are  CD,  DE,  and  EA,  and  their  resultant  in  magnitude 
and  direction  is  ca,  which  is  opposed  by  the  equal  and  opposite 
resultant  ac  of  the  stresses  in  the  members  A  and  C.  It  is  thus 
seen  that  on  either  side  of  the  section  xy  there  is  equilibrium 
between  the  external  forces  and  the  internal  stresses  in  the  mem- 
bers cut  by  xy,  and  on  this  principle  is  founded  the  section 
method  of  determining  stresses,  to  be  referred  to  later. 

A  practical  application  of  the  funicular  polygon  will  be  made 
by  taking  the  beam  of  Fig.  27,  p.  31,  reproduced  in  Fig.  84. 
The  linear  scale  is  i  inch  =  4  feet,  and  the  load  scale  i  inch  = 
150  pounds.  Then,  W\  =  60  pounds  =  0.4  inch  to  scale, 
Wz  =  45  pounds  =  0.3  inch,  and  Ws  =  90  pounds  =  0.6  inch. 
Letter  the  beam  according  to  the  Bow  system,  so  that  Wi,  W2, 
W,,  R2,  and  R!  will  be  known  as  AB,  BC,  CD,  DE,  and  EA 
respectively. 

To  construct  the  force  diagram  we  set  off,  vertically  downward, 
ab  equal  in  length  to  0.4  inch  to  represent  the  downward  force  Wi, 
or  AB,  to  scale;  be  equal  in  length  to  0.3  inch  to  represent  W2, 
or  BC;  and  cd  equal  in  length  to  0.6  inch  to  represent  TF3,  or 
CD.  Then  we  know  that  da  is  the  closing  line  of  the  force 


FUNICULAR   POLYGON 


123 


diagram,  all  the  forces  being  vertical,  and  that  it  represents 
the  sum  of  the  reactions  RI  and  R^\  but  we  do  not  know  the 
amount  of  the  load  borne  by  each  support. 


w., 


a     B 


k 

RI 

E 

1  1 
H 

A 

1  ' 

|  I 

1  1 

R=ce 
A 


Fig.  84. 

To  find  RI  and  R%  we  must  construct  the  funicular  polygon 
of  the  forces.  Select  at  random  some  point  o,  distant  oh  from 
the  load  line  ab,  as  a  pole  and  draw  the  vectors  oa,  ob,  oc,  and  od. 
From  some  point  j  in  the  line  of  action  of  RI  draw  a  line  parallel 
to  the  vector  oa,  and  from  its  point  of  intersection,  k,  with  the 
vertical  from  W\  draw  a  line  parallel  to  the  vector  ob,  and  from 
its  point  of  intersection,  /,  with  the  line  of  action  of  W2  draw 
a  line  parallel  to  vector  oc,  and  from  its  point  of  intersection,  m, 


124          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

with  the  line  of  action  of  W$  draw  a  line  parallel  to  the  vector 
od.  This  last  line  intersects  the  line  of  action  of  R%  at  n.  Join 
n  with  jj  and  we  have  the  funicular  polygon  jkfmn  with  the 
external  forces  of  the  beam  acting  at  its  joints.  This  funicular 
polygon  has  five  sides,  while  there  are  but  four  vectors  in  the 
force  polygon,  and  since  there  must  be  a  vector  for  each  side 
of  the  funicular  polygon,  the  vector  oe,  parallel  to  njt  must  be 
drawn.  The  point  e  is  thus  determined,  and  de  represents  in 
magnitude  and  direction  the  support  reaction  DE  =  R2  to  the 
scale  adopted;  and  in  like  manner  ea  represents  in  magnitude 
and  direction  the  support  reaction  EA  =  RI.  By  measurement 
de  is  M  inch,  which,  reduced  to  scale,  equals  fj  X  150  =  127.5 
pounds  =  R2;  and  ea  measures  f$  inch,  which,  reduced  to  scale, 
is  67.5  pounds  =  RI.  These  are  the  same  values  found  before 
for  RI  and  R2. 

It  will  not  be  necessary  to  letter  the  funicular  polygon  in  order 
to  know  the  names  of  the  members,  as  an  examination  of  the 
force  polygon  at  once  discloses  them.  Thus,  the  member 
parallel  to  oa  is  OA,  the  one  parallel  to  ob  is  OB,  and  so  on. 
The  names  of  the  members  may  also  be  determined  from  the 
lettering  of  the  beam,  since  the  length  of  each  member  is  termi- 
nated by  the  lines  of  action  of  two  forces  of  the  beam.  For 
example,  the  member  kf  is  terminated  by  the  lines  of  action  of 
Wi  and  W2,  and  since  the  letter  B  appears  between  W\  and  Wz 
the  member  kf  is  named  OB,  or  simply  B. 

The  resultant,  ad,oi  Wi,  W2,  and  Ws  acts  through  r  (see  Art.  57). 
Should  it  be  desired  to  replace  two  or  more  of  the  forces  by  their 
resultant,  its  magnitude  and  a  point  in  its  line  of  application 
may  be  determined.  Thus,  if  it  were  desired  to  replace  the 
forces  Wz  and  R2  by  their  equivalent,  we  find  from  the  force 
diagram  that  the  resultant  of  cd  and  de  is  ce  in  magnitude  and 
direction,  and  a  point  in  the  line  of  application  is  found  by 
producing  OE  and  OC  to  their  intersection  as  shown. 


FUNICULAR  POLYGON  12$ 

59.   The  Funicular  Polygon  a  Bending-moment  Diagram.  - 

Consider  any  section  a  of  the  beam  of  Fig.  84.  Produce  jk  to 
its  intersection  with  the  vertical  from  a  at  p.  Draw  the  hori- 
zontals x  and  xi  and  regard  them  as  the  altitudes  of  the  triangles 
jpz  and  kpy  respectively.  All  the  triangles  of  the  force  polygon 
have  the  same  altitude  oh. 

The  triangles  jpz  and  kpy  are  respectively  similar  to  the 
triangles  oea  and  oba,  having  their  sides  mutually  parallel. 
Hence  we  have 

-^-  =  —  =  2-  j   whence    RIX  =  pz  X  oh, 
oh      ea      RI 


and  7  =       =      -  ,   whence   W&i  =  py  X  oh. 

oh      ab      Wi 

The  bending  moment  at  the  section  a  is, 

Ma  =  RIX  —  WiXi  =  pz  X  oh  —  py  X  oh  =  oh  (pz  —  py)  =  yzX  oh. 

That  is,  the  bending  moment  at  any  section  of  the  beam  is  equal 
to  the  product  of  the  ordinate  of  the  funicular  polygon  at  the 
section  and  the  polar  distance.  Thus,  the  ordinate  yz  under  the 
section  a  measures  ft  inch,  and  the  polar  distance  oh  measures 
i  inch,  representing  150  pounds  to  the  scale  selected.  Then  the 
bending  moment  at  section  a  is, 

Ma  =  f  f  X  4  X  150  =  217.5  pounds-feet, 

as  was  found  on  page  33. 

The  shear  diagram  of  the  beam  can  readily  be  constructed 
by  projection  from  the  force  diagram.  Thus,  the  shear  at  any 
section  between  the  left  support  and  W\  is  RI  =  ea,  and  is 
plotted  by  projecting  ea  horizontally  as  shown.  At  any  section 
between  Wi  and  W2  the  shear  is  .Ri  —  W\  =  ea  —  ab  =  eb,  and 
is  plotted  by  projecting  eb.  The  shear  at  any  section  between 
W2  and  Ws  is  RI  —  W\  —  W2  =  ea  —  ab  —  be  =  —  ec,  and  be- 
tween WB  and  the  right  support  the  shear  is  RI  —  W\  —  Wz  —  Ws 


126          THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 

=  ea  —  ab  —  be  —  cd  =  —  ed  =  —  R%.  Projecting  these  two 
shears,  the  diagram  is  completed. 

It  has  been  stated  that  the  pole  of  the  force  diagram  may  be 
selected  at  random,  but  if  it  be  selected  so  that  its  distance  from 
the  load  line  be  some  definite  number  expressed  to  scale  in  units 
of  the  load,  then  a  bending-moment  scale  may  be  obtained  which 
will  enable  the  bending  moment  to  be  measured  directly  from 
the  diagram  and  obviate  the  necessity  of  multiplying  each 
measurement  by  the  polar  distance.  The  pole,  0,  of  the  force 
diagram  of  Fig.  84  was  taken  at  a  distance  of  i  inch  from  the 
load  line  ab,  the  polar  distance  oh,  therefore,  representing 
150  pounds.  Then,  since  the  linear  scale  is  i  inch  =  4  feet,  an 
ordinate  measuring  i  inch  represents  4  feet;  but  as  this  must  be 
multiplied  by  the  polar  distance  we  shall  have: 

i  inch  =  4  feet  X  150  pounds  =  600  pounds-feet, 
or         sV  inch  =  10  pounds-feet, 

a  new  and  convenient  scale  by  which  the  bending  moments  can 
be  measured  directly  from  the  diagram.  The  bending-moment 
scale  is  derived  in  each  instance  by  multiplying  the  linear  scale 
by  the  polar  distance  expressed  in  pounds  or  tons.  The  ordinate 
yz  of  the  funicular  polygon,  Fig.  84,  measures  21.75  sixtieths  of 
an  inch,  and  the  bending  moment  at  the  section  a  of  the  beam  is, 
therefore,  21.75  X  10  =  217.5  pounds-feet. 

A  few  examples  of  the  application  of  the  funicular  polygon  to 
beams  will  be  given. 

Example  I.  —  A  beam  supported  at  the  ends  is  20  feet  long, 
weighs  400  pounds,  and  has  concentrated  loads  of  360  pounds 
and  440  pounds  at  8  feet  from  the  left  end  and  4  feet  from  the 
right  end  respectively.  Draw  the  bending-moment  and  shear 
diagrams,  and  measure  the  bending  moments  under  the  con- 
centrated loads  and  the  shear  stress  at  the  middle  of  the 
beam. 


FUNICULAR  POLYGON  127 

Solution.  —  Select  a  linear  scale  of  J  inch  =  i  foot,  and  a  load 
scale  of  i  inch  =  400  pounds. 

Set  out  the  beam  as  shown  in  Fig.  85,  and  since  the  beam 
weighs  400  pounds  it  has,  in  addition  to  the  concentrated  loads, 
a  uniformly  distributed  load  of  20  pounds  per  foot. 

We  shall  first  construct  the  funicular  polygon  for  the  concen- 
trated loads,  neglecting  for  the  present  the  uniformly  distributed 
load. 

Set  off  the  load  line  ac  by  making  ab  measure  0.9  inch  and  be 
measure  i.i  inches  to  represent  the  loads  of  360  pounds  and 
440  pounds  respectively.  Select  the  pole  o  at  a  distance  of  1.25 
inches  from  the  load  line,  so  that  the  polar  distance  will  represent 
1.25  X  400  =  500  pounds.  We  shall  then  have  for  the  bending- 
moment  scale,  J  inch  =  i  foot  X  500  pounds  =  500  pounds-feet, 
or  ?V  inch  =  100  pounds-feet. 

Draw  the  vectors,  oa,  ob,  and  oc.  From  a  point  s  in  the  line  of 
action  of  RI  draw  a  parallel  to  oa  and  produce  it  until  it  intersects 
the  line  of  action  of  W\  at  the  point  r.  From  r  draw  a  parallel 
to  ob,  producing  it  until  it  intersects  the  line  of  action  of  Wi  at 
the  point  j.  From  j  draw  a  line  parallel  to  oc  and  produce  it 
until  it  intersects  the  line  of  action  of  R2  at  the  point  k.  Join 
k  with  s.  Then  js  is  the  closing  line  of  the  funicular  polygon 
srjk.  Draw  od  parallel  to  sk.  Then  cd  and  da  represent  in 
magnitude  and  direction  the  support  reactions  R%  and  RI  re- 
spectively, and  the  funicular  polygon  srjk  is  the  diagram  of 
bending  moments  for  the  concentrated  loads. 

The  bending-moment  diagram  of  the  evenly  distributed  load 
of  400  pounds  will  be  parabolic  in  form,  and  it  will  be  convenient 
to  construct  its  funicular  on  the  closing  line  ks  of  the  funicular 
of  the  concentrated  loads.  To  do  so  we  will  take  the  same  pole, 
o,  as  was  used  for  the  concentrated  loads,  and  will  consider  the 
whole  of  the  distributed  load  of  400  pounds  to  be  concentrated 
at  the  middle  of  the  beam,  as  shown  in  the  figure.  This  will  add 


128  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 


(F)      (400  Ibs.)    (E) 


.  Fig.  85. 


FUNICULAR  POLYGON  129 

200  pounds  each  to  R2  and  RI.  On  each  side  of  d  in  the  load 
line  lay  off  ed  and  df,  each  one-half  inch  in  length,  to  represent 
these  additions  to  the  support  reactions.  Draw  the  vectors 
oe  and  of.  From  k  and  5  draw  parallels  to  oe  and  of  respectively. 
They  intersect  at  z.  Then  zsk  is  the  funicular  polygon  for  the 
distributed  load,  supposing  it  to  be  concentrated  at  the  middle 
of  the  beam.  It  has  been  shown  in  Art.  18  that  the  bending 
moment  at  the  middle  of  a  simple  beam  uniformly  loaded  is  only 
one-half  that  due  to  the  same  load  concentrated  at  the  middle. 
Hence,  the  ordinate  ti  measures  the  bending  moment  at  the 
middle  of  the  beam  due  to  the  distributed  load,  i  being  the  middle 
point  of  tz.  A  parabola  constructed  on  sk  as  a  chord,  and  passing 
through  the  point  2,  is  the  bending-moment  diagram  due  to  the 
distributed  load,  and  srjki  is  the  complete  bending-moment 
diagram  for  the  beam. 

The  reaction  R2  measured  on  the  load  line  is  cd  +  df  =  H 
inches;  hence,  R2  =  H  X  400  =  696  pounds.  The  reaction 
RI  =  da  H-  de  =  H  inches;  hence,  RI  =  f g-  X  400  =  504  pounds. 

The  ordinates  y  and  y'  under  Wi  and  W2  measure  f$  inch  and  |$ 
inch  respectively.  The  bending  moments  under  W\  and  W2  are 
therefore  3400  pounds-feet  and  2600  pounds-feet  respectively. 
These  results  may  be  checked  easily  by  calculation. 

Commencing  at  the  left  support  the  shear  due  to  the  uniform 
load  is  equal  to  half  that  load,  and  gradually  decreases  from  left 
to  right  until,  at  the  middle,  it  becomes  zero,  and  at  the  right 

support  it  becomes wL  = >  or  to  one-half  the  load, 

2  2 

but  negative,  w  denoting  the  load  per  unit  of  length.  The  dia- 
gram d'e'fd",  therefore,  represents  the  shear  due  to  the 
distributed  load. 

The  total  shear  at  the  left  support  is  RI  and  is  equal  to  da  +  de. 
The  parallel  to  e' ' f  shows  the  gradual  decrease  of  the  shear  from 
the  left  support  to  W\  due  to  the  uniform  load.  Passing  W\ 


136         THE  ELEMENTS  OF  MECHANICS   OF   MATERIALS 

the  shear  suddenly  drops  to  n  and  becomes  negative,  mn  being 
equal  to  db.  The  parallel  to  e* f  drawn  from  n  shows  the  gradual 
negative  increase  in  the  shear  from  Wi  to  W2  due  to  the  dis- 
tributed load.  Passing  W2  the  shear  suddenly  drops  to  q,  pq 
being  equal  to  be.  The  parallel  to  e' ' jr  drawn  from  q  shows  the 
further  increase  in  the  shear  due  to  the  distributed  load  until, 
at  Vj  it  becomes  —  R2  =  —  (dc  +  df).  The  ordinate  uw  at  the 
middle  of  the  beam  measures  /o  inch;  the  shear  at  the  middle 
section  is,  therefore,  /o  X  400  =  56  pounds. 

Example  II.  —  The  beam  with  overhanging  ends  of  Fig.  52, 
p.  52,  loaded  uniformly  with  20  pounds  per  foot  and  with  two 
concentrated  loads,  may  be  solved  readily  by  means  of  the 
funicular  polygon.  The  concentrated  loads  W\  and  W%  are 
200  pounds  and  400  pounds  respectively. 

Set  off  the  beam  as  shown  in  Fig.  86.  Divide  the  beam  into 
any  convenient  number  of  parts,  eight  in  this  instance,  so  that 
each  part  will  be  4  feet  in  length  and  will  bear  80  pounds  of  the 
distributed  load.  These  eight  parts  constitute  as  many  external 
forces,  each  acting  at  its  center  of  gravity. 

Letter  the  beam  according  to  the  Bow  system,  and  adopt  the 
following  scales:  Linear,  o.i  inch  =  i  foot;  load,  i  inch  =  400 
pounds.  We  shall  then  have:  W\  =  f£§  -  0.5  inch;  W2  =  -f§# 
=  i  inch;  and  the  load  of  each  of  the  equal  divisions  will  be 
represented  by  ^o  =0.2  inch. 

Set  off  the  load  line  ak  accordingly.  Select  a  pole  o  distant 
1.5  inches  from  the  load  line,  so  that  the  polar  distance  will 
represent  600  pounds.  We  shall  then  have:  TV  inch  =  i  foot  X 
600  =  600  pounds-feet,  or  ^  inch  =  100  pounds-feet,  for  the 
bending-moment  scale. 

Draw  the  vectors  of  the  force  polygon.  From  some  point  m 
in  the  line  of  action  of  RI  draw  a  parallel  to  oa  and  produce  it 
until  it  intersects  the  line  of  action  of  AB  at  n.  Through  n 
draw  a  parallel  to  ob,  producing  it  to  its  intersection  p  with 


FUNICULAR   POLYGON 


200  Ibs. 


Fig.  86. 


132          THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 

the  line  of  action  of  BC,  or  of  W\.  Draw  parallels  to  the  other 
vectors  as  shown.  The  member  qr,  parallel  to  07,  intersects  the 
line  of  action  of  JK  at  r.  The  parallel  to  ok  through  r  intersects 
the  line  of  action  of  R2  at  s.  The  closing  member  is,  then,  sm, 
and  mnp  .  .  .  qrs  is  the  funicular  polygon,  or  the  bending- 
moment  diagram  of  the  beam. 

Draw  ol  parallel  to  sm.  Then  kl  and  la  represent  to  scale  the 
reactions  R2  and  RI  respectively. 

It  should  be  noted  that  ordinates  within  mnpt  and  sur  give 
negative  bending  moments,  and  that  /  and  u  are  the  points  of 
inflection.  The  ordinates  y  and  y'  give  the  maximum  negative 
and  positive  bending  moments  respectively,  and  as  they  measure 

-  and  —  of  an  inch  respectively,  the  bending  moments  are 

DO  OO 

1440  pounds-feet  and  1600  pounds-feet. 

Commencing  at  the  left  end  of  the  beam,  the  shear  increases 
from  zero  to  —  ab  when  W\  is  reached.  Passing  W\  the  shear 
becomes  —  ab  —  be  =  —  ac,  and  increases  up  to  the  left  sup- 
port, where  it  becomes  —  (ab  +  be  +  cd)  =  —  ad.  Passing  the 
left  support  it  becomes  RI  —  ad  =  Id  and  decreases  up  to 
Wz,  where  it  becomes  Id  —  dg  =  Ig.  Passing  Wz  it  becomes 
lg  —  gh  =  —  Ikj  and  gradually  increases  until  the  right  support 
is  reached,  where  it  becomes  —  Ih  —  hz  =  —  Iz  (the  center  of 
gravity  of  the  weight  on  the  seventh  division  of  the  beam  happen- 
ing to  fall  directly  over  the  right  support,  one-half  of  that  weight 
(//)  actually  lies  to  the  left  and  one-half  to  the  right  of  the 
support,  and  must  be  so  considered).  Passing  the  right  support 
the  shear  again  becomes  positive  and  equal  to  R%  —  Iz  =  zk,  and 
then  decreases  to  zero  at  the  right  end  of  the  beam. 

Example  ILL  —  To  draw  the  bending-moment  and  shear  dia- 
grams of  the  cantilever  with  concentrated  loads,  as  shown  in 
Fig.  87,  we  proceed  as  follows: 

Draw  the  load  line  ad,  making  ab,  be  and  cd  equal,  to  some 


FUNICULAR   POLYGON 


133 


selected  scale,  to  the  loads  A  B,  BC,  and  CD  respectively.  Select 
some  pole  o,  and  draw  the  vectors  oa,  ob,  oc,  and  od.  From  some 
point  r  in  the  support  line  draw  a  parallel  to  oa,  producing  it 
until  it  intersects  the  line  of  action  of  AB  at  q.  From  q  draw  a 
parallel  to  ob  and  produce  it  until  it  intersects  the  line  of  action 
of  BC  at  p.  From  p  draw  a  parallel  to  oc  and  produce  it  until 
it  intersects  the  line  of  action  of  CD  at  n.  From  n  draw  a 
parallel  to  od  to  meet  the  line  of  support  at  m.  Then  mnpqr  is 


Fig.  87- 

the  bending-moment  diagram  of  the  cantilever,  and  the  ordi- 
nate  under  any  section  of  the  beam  is  the  measure  of  the  bending 
momejit  at  that  section.  The  maximum  bending  moment  is, 
of  course,  at  the  wall. 

The  shear  diagram  is  drawn  by  projection  from  the  load  line 
and  presents  no  difficulties. 

In  the  case  of  a  cantilever  it  is  found  convenient  to  select  the. 
pole  at  some  chosen  perpendicular  distance  from  either  extrem- 


134 


THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 


ity  of  the  load  line.  In  Fig.  87  the  pole  o  might  just  as  well  have 
been  chosen  at  o'  and  an  equal  bending-moment  diagram, 
mn'p'q'r',  constructed,  as  shown  by  the  dotted  lines. 

Example  IV.  —  The  cantilever  of  Fig.  39,  p.  40,  having  a  com- 
bination of  concentrated  and  uniformly  distributed  loads  may  be 
solved  easily  by  means  of  the  funicular  polygon. 

Using  the  same  linear  scale  of  T2o  inch  =  i  foot  as  for  Fig.  39, 
and  a  load  scale  of  400  pounds  to  the  inch,  we  shall  first  con- 
struct the  funicular  polygon  for  the  concentrated  loads. 

Reduced  to  scale,  AB  =  gfa  =  Tf ^  inch,  and  BC  =  ¥V&  = 
TVo  inch.  Set  off  the  load  line  ac}  Fig.  88,  making  ab  and  be 
equal  to  Tf  Q  inch  and  TW  inch  respectively. 


(D)(192  Ibs.) 


32  Iba. 


Fig.  88. 

Choose  a  pole  o  at  a  perpendicular  distance  of  i  inch  from  a, 
so  that  the  polar  distance  oa  represents  400  pounds.  We  shall 
then  have  for  the  bending-moment  scale,  T2o  inch  =  i  foot  X 
400  pounds  =  400  pounds-feet,  whence  -fa  inch  =  40  pounds-feet. 
Draw  the  vectors  oa,  ob,  and  oc. 


FUNICULAR   POLYGON  135 

Commencing  at  some  point  q  in  the  wall  line,  draw  qp  parallel 
to  oa,  pn  parallel  to  ob,  and  nm  parallel  to  oc,  thus  forming  the 
funicular  polygon,  or  bending-moment  diagram,  mnpq  for  the 
concentrated  loads. 

The  cantilever  is  uniformly  loaded  with  48  pounds  per  foot 
for  a  distance  of  4  feet  from  the  wall,  making  a  total  uniform 
load  of  192  pounds.  Assuming  this  load  concentrated  at  the 
outer  extremity  of  the  4  feet,  set  off  ad  equal  to  J$§  =  ff  inch 
to  represent  it  (the  uniform  load  AD  is  taken  contraclockwise 
in  order  to  join  its  funicular  to  the  line  qp  of  the  funicular  poly- 
gon of  the  concentrated  loads).  Draw  the  vector  od,  and  from 
r,  the  intersection  of  pq  with  the  vertical  at  4  feet  from  the  wall, 
draw  rt  parallel  to  od.  It  can  easily  be  shown  that  the  bending 
moment  due  to  a  concentrated  load  at  the  end  of  a  cantilever  is 
twice  that  due  to  the  same  load  uniformly  distributed,  so  the 
distance  qt  must  be  bisected  at  s,  and  the  parabola  having  its 
apex  at  r,  and  passing  through  s,  gives  rsq  as  the  bending-moment 
diagram  due  to  the  distributed  load.  Then  mnprs  is  the  com- 
plete bending-moment  diagram  of  the  cantilever.  The  ordinate 


2  CJ   'Z 

ms  at  the  wall  measures  -^  inch,  and  the  bending  moment  is, 

5° 

therefore,  25.3  X  40  =  1012  pounds-feet,  the  same  as  found  on 
page  41. 

The  shear  diagram  is  drawn  by  projection  from  the  load  line 
and  presents  no  difficulties. 

Example  V.  —  A  circular  steel  axle,  supported  on  end  jour- 
nals, is  10  feet  long  between  journal  centers  and  subjected  to  a 
load  of  12,000  pounds  at  a  point  4  feet  from  the  center  of  the 
journal  at  the  right  end.  Determine  the  dimensions  of  the 
axle. 

Solution.  —  The  axle  is  to  be  regarded  as  being  subjected  only 
to  a  bending  stress. 

To  a  scale  of  f  inch  =  i  foot,  lay  off  the  axle  in  skeleton  form, 


136          THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 

Fig.  89.  To  a  load  scale  of  f  inch  =  1200  pounds,  lay  off  the 
load  line  abj  1.25  inches  in  length,  to  represent  the  load  of  12,000 
pounds.  Select  a  pole  o  one  inch  from  ab,  so  that  the  polar 


distance  oh  represents  9600  pounds.  Then,  f  inch  =  i  foot 
X  9600  pounds  =  9600  pounds-feet,  or  ?V  inch  =  640  pounds- 
feet,  a  convenient  bending-moment  scale. 


FUNICULAR   POLYGON  137 

Draw  the  vectors  oa  and  ob.  Commencing  at  a  point  m  in 
the  line  of  action  of  RI,  draw  the  funicular  polygon  mjk.  Draw 
oc  parallel  to  the  closing  line  km.  Then  be  and  ca  are  the  re- 
actions RI  and  RI  respectively,  in  magnitude  and  direction.  By 
scale  measurement  be  =  7200  pounds  =  R^  and  ca  measures 
4800  pounds  =  RI. 

Treating  the  journals  as  uniformly  loaded  cantilevers,  we  have 

7?  7 
—  as  the  maximum  bending  moment  of  the  journal  at  the  left 

end,  in  which  I  denotes  the  length  of  the  journal. 


™,          .    .  .  SI  e  ^j,  .       ,  .  , 

The  resisting  moment  is  —  =  -  —  =  --  =  0.196  Sd3,  in  which 

c        64  c       32 

d  denotes  the  diameter  of  the  journal.  From  this  it  is  seen  that 
the  resisting  moment,  and  therefore  the  bending  moment,  is  pro- 
portional to  the  cube  of  the  diameter. 

Then,         ^  =  0.196  Sd8,    whence    d  =  \J     R*-^' 
2  Y  0.1960    d 

It  is  usual  to  fix  the  ratio  —  in  accordance  with  the  conditions 

d 

of  the  case.  Assuming  the  direction  of  the  load  to  be  constant 
and  the  maximum  revolutions  to  be  250  per  minute,  it  is  good 

practice  to  make  -  :  =  2  ;  for  slower  speeds  the  ratio  would  be  less. 
d 

Then,  taking  S  at  10,000  pounds  per  square  inch,  we  have 


d  =  0.0226  A/48oo  X  2  =  2.21  inches,  say  2\  inches; 
whence  /  =  4^  inches. 

For  the  journal  at  the  right  end, 


df=  0.0226  Vj2oo  X  2  =  2.71  inches,  say  2\  inches;  and 
I'  =  5.5  inches. 

Since  the  bending  moment  at  any  section  of  the  axle  is  pro- 
portional to  the  cube  of  the  diameter  at  the  section,  it  follows 
that  the  diameter  at  any  section  is  proportional  to  the  cube  root 


138          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

of  the  bending  moment  at  the  section.  The  lengths  I  and  /' 
being  known,  the  bending  moments  M  and  M'  may  be  measured 

7?  7          7?  /' 
from  the  diagram  or  calculated  from  — —  and  — —  respectively. 

These  moments  are  found  to  be:  M  =  625  pounds-feet,  and 
M'  =  1200  pounds-feet.  The  bending  moment  MI  at  the  load 
measures  28,800  pounds-feet. 


Then,  §--*--  \7~J  =  V  -  2. 

d       2.71       v  M  1200 

whence         di  =  7.8  inches,  say  7!!  inches. 

Making  the  hub  seat  of  the  wheel  8  inches  long,  we  have  now 
to  determine  d%  and  d^  The  bending  moments  M2  and  M3 
measure  26,240  pounds-feet  and  27,200  pounds-feet  respectively. 


r™  d^          d>  *  /26,240 

Then,  -  = =  V  -        -  =  2.796, 

d       2.71  1200 

whence  d?  =  7.577  inches,  say  7!  inches. 


dz        dz        .8/27,2oo 

and  -~  = =  V  -k-  -  =  3.517, 

(/       2.21       V     625 

whence  J3  =  7.773  inches,  say  7!  inches. 

An  empirical  rule  in  machine  design  makes  the  height  of  the 

collar  of  the  journal —  +  -inch,  and  its  width  1.5  times  the  height. 
10     8 

Hence, 

Height  of  journal  collar  at  left  end  =  — h  -  =  —  inches, 

10       8      32 

whence  width  =  ^  inches. 

Height  of  journal  collar  at  right  end  =  —^  +  0  =  0  inches, 

10       8      8 

whence  width  =  T\  inch. 

The  axle  is  drawn  in  Fig.  90  to  the  scale  of  one-sixth. 
Example  VI.  —  The  total  weight  on  the  4  axles  of  a  semicon- 
vertible  street  car  is   46,560  pounds.     Motion  is  conveyed  to 


FUNICULAR    POLYGON  139 

the  axles  from  the  motors,  one  for  each  axle,  by  means  of 
gears,  and  the  weight  is  carried  on  journals  that  overhang  the 
car- wheel  journals  9.25  inches.  The  horse  power  of  each  motor 
is  40,  and  the  maximum  speed  on  level  is  25  miles  per  hour.  The 
diameter  of  the  car  wheels  is  33  inches  and  the  gauge  of  the  road 
5  feet.  Determine  the  dimensions  of  the  axles. 

Solution.  —  A  car  axle  is  subjected  to  bending  only  when  the 
car  is  moving  on  a  straight  portion  of  the  road,  but  in  rounding 
curves  it  is  subjected  to  both  torsion  and  bending,  so  that  the 
equivalent  twisting  moment  must  form  the  basis  for  the  deter- 
mination of  the  dimensions  of  the  axle. 

The  load  carried  on  each  journal  is  — -  =  ^-~ —  =  5820  pounds, 

8  8 

which  occasions  an  equal  reaction  at  the  journal  of  each  car 
wheel.  The  condition  then  is  one  of  a  beam  having  equal  over- 

W 

hanging  ends,  each  overhang  being  uniformly  loaded  with  —  or, 

8 

W 
which  is  the  same  thing,  having  —  concentrated  at  its  middle. 

8 

To  a  linear  scale  of  f  inch  =  i  foot,  or  of  one- twentieth,  lay  off 
the  axle  in  skeleton  form,  Fig.  91.  To  a  load  scale  of  i  inch  =  582 
pounds  lay  off  ab  1.25  inches  in  length  to  represent  the  load  AB 
of  5820  pounds,  and  be  the  same  length  to  represent  the  equal 
load  BC.  Select  a  pole  0,  distant  1.25  inches  from  the  load 
line  ac,  so  that  the  polar  distance  will  represent  5820  pounds. 
Then,  f  inch  =  i  foot  X  5820  pounds  =  5820  pounds-feet,  or  3*0 
inch  =  194  pounds-feet,  which  is  a  convenient  bending-moment 
scale. 

Draw  the  vectors  oa,  ob,  and  oc.  Commencing  at  some  point 
m  in  the  line  of  action  of  RI,  draw  ma  parallel  to  oa,  and  from  a, 
the  point  of  intersection  of  ma  with  the  line  of  action  of  AB, 
draw  an  parallel  to  ob,  and  from  n,  its  intersection  with  the  line 
of  action  of  BC,  draw  nk  parallel  to  oc,  intersecting  the  line  of 
action  of  R%  at  k.  Join  k  with  m.  Then  mank  is  the  funicular 


140          THE  ELEMENTS   OF  MECHANICS  OF   MATERIALS 


Fig.  92. 


FUNICULAR  POLYGON 

polygon,  or  b ending-moment  diagram,  of  the  axle.     Draw  od 
parallel  to  km\  it  will,  of  course,  coincide  with  ob,  since  ab  and 
be  are  equal  and  o  was  taken  on  the  horizontal  through  b. 
The  bending  moment  M  between  the  supports  is  constant  and 


measures          X  194  =  4500  pounds-feet. 

To  find  the  twisting  moment  we  must  know  the  revolutions 
per  minute  of  the  axle. 

Circumference  of  car  wheel  =  ^—^  -  ^  =  8.64  feet. 

12 

Speed  of  car  per  minute  =  -  —  -  -  ^  =  2200  feet. 

2200  , 

r'p'm-  =  s^T  =  254'63* 

Pr  =M<  =         40  X  33.000          =  ds.feet> 

2  X  3.1416  X  254.63 

To  find  the  equivalent  twisting  moment  (see  Art.  47)  we  have 

2 


Me  =  M  +  VM2+M*  =  4500  +  V(4Soo)2  +  (825) 
=  9075  pounds-feet  =  108,900  pounds-inches. 

Taking  S  as  6000  (see  Art.  46),  we  shall  have 

108,900  =  0.196  Sd?  =  0.196  X  6000  d3, 
in  which  d  is  the  diameter  of  the  axle. 

r™  ,  3/I08,900  .          ,  ,     .         , 

Then          d  =  y  -    ^—-  =  4.52  inches,  say  4i  inches. 

Treating  the  overhanging  journals  as  cantilevers  uniformly 

loaded,  or  with  the  load  -  -  concentrated  at  the  middle,  the 

8 

maximum  bending  moment  is  M  =  4500  pounds-feet.     For  the 
diameter  of  the  journals  we  shall  then  have 


^  /  C  A    OOO 

4500 X  12  =  0.196 Sdi3,  whence  di=  y  =3-58  inches. 


142         THE   ELEMENTS   OF  MECHANICS  OF  MATERIALS 

Or,  since  the  bending  moments  are  proportional  to  the  cubes 
of  the  diameters,  we  shall  have,  denoting  the  diameter  of  the 
journals  by  dit 


\7  ^-^  =  1.26,    whence   d\  =  3.59  inches,  say  3!  inches. 

V      °° 


~ 

45°° 

Denoting  the  length  of  the  journals  by  /,  and  taking  the  ratio 

—-as  2,  we  shall  have 
di 

Length  of  journal  =  3.59  X  2  =  7.18  inches,  say  77%  inches. 

The  height  of  collar  at  outer  end  of  journal  =  ^^  +  -  = 

10        8 

0.484  inch,  say  J  inch. 

Width  of  collar  =  0.484  X  1.5  =  0.7  inch,  say  tJ  inch. 

The  diameter  of  the  axle  has  been  determined  under  the 
supposition  that  the  gear  wheel  which  receives  the  power  from 
the  motor  is  fitted  to  the  axle  by  hydraulic  pressure.  Should 
the  gear  wheel  be  keyed  to  the  axle  the  diameter  would  have  to 
be  increased  by  an  amount  equal  to  the  depth  of  the  key  way, 
amounting  in  this  instance  to  about  fV  inch. 

The  axle  is  drawn  in  Fig.  92  to  the  scale  of  one-sixth. 

Example  VII.  —  A  steel  axle  rests  on  two  journals  and  is 
subjected  to  a  vertical  load  of  6000  pounds  applied  on  an  over- 
hanging end  at  a  distance  of  15  inches  from  the  center  of  the 
nearest  journal.  The  distance  between  the  journals  is  4  feet. 
Construct  the  funicular  polygon,  and  determine  the  magnitude 
and  direction  of  the  reactions. 

Solution.  —  To  a  linear  scale  of  J  inch  =  i  foot  lay  off  the 
axle  in  skeleton  form,  Fig.  93.  To  a  load  scale  of  J  inch  = 
1500  pounds  lay  off  the  load  line  ab  one  inch  in  length  to  repre- 
sent the  load  AB  of  6000  pounds.  From  the  pole  o,  distant  i 
inch  from  ab,  draw  the  vectors  oa  and  ob.  The  polar  distance 
of  i  inch  represents  6000  pounds  to  scale,  and  we  shall  have 

Jinch  =  i  ft.  X  6000  Ibs.  =  6ooolbs.-ft.,  or  *V  inch  =  2oolbs.-ft. 
as  a  bending-moment  scale. 


FUNICULAR   POLYGON 


143 


From  a  point  m  in  the  line  of  action  of  RI  draw  a  parallel  to 
ad,  intersecting  the  line  of  action  of  W  at  j.  From  j  draw  a 
parallel  to  ob,  intersecting  the  line  of  action  of  R%  at  k,  and  from 
k  draw  km  as  the  closing  line  of  the  funicular  polygon  mjk. 
From  o  draw  a  parallel  to  km,  intersecting  the  load  line  pro- 
duced at  c.  Then  be  and  ca  represent  the  reactions  R2  and  RI 
in  magnitude  and  direction  respectively.  By  scale  measurement 
be  =  RI  —  7875  pounds,  and  ca  =  RI  =  1875  pounds. 


m 


7:7; 


Fig.  93- 


PROBLEMS 

1.  A  beam  22  feet  long  supports  a  load  of  1000  pounds  at  a  point  6 
feet  from  the  left  end  and  one  of  1200  pounds  at  5  feet  from  the  right  end. 
In  addition  there  is  a  uniformly  distributed  load  of  100  pounds  per  foot. 
Construct  the  bending-moment  and  shear  diagrams.     What  concentrated 
load  must  be  placed  at  the  middle  of  a  similar  beam  in  order  that  the 
maximum  bending  moment  shall  be  the  same  as  that  of  the  given  beam  ? 

Ans.    2190  Ibs. 

2.  A  beam  30  feet  long  weighs  20  pounds  per  foot  and  overhangs  each 
support  6  feet.     It  bears  a  superimposed  load  of  100  pounds  per  foot,  and 
a  load  of  1400  pounds  concentrated  at  a  point  3  feet  to  the  right  of  the 


144          THE   ELEMENTS   OF  MECHANICS  OF    MATERIALS 

middle.  Construct  the  bending-moment  and  shear  diagrams,  and  find 
graphically  the  bending  moment  at  the  dangerous  section  and  the  dis- 
tances of  the  points  of  inflection  from  the  left  support. 

Ans.    7760  lbs.-ft.;  1.48  ft.  and  16.9  ft. 

3.  A  cantilever  14  feet  long  supports  three  concentrated  loads,  —  500 
pounds  at  4  feet  from  the  wall,  600  pounds  at  10  feet  from  the  wall,  and 
200  pounds  at  the  extremity.     In  addition  it  bears  a  uniformly  distrib- 
uted load  of  80  pounds  per  foot  run.     Construct  the  bending-moment  and 
shear  diagrams. 

4.  Determine  the  diameters  of  the  journals  of  the  axle  of  Example  VII, 
page  142.  Ans.    if  ins.  and  4^  ins. 


CHAPTER  VIII 
FRAMED   STRUCTURES.     RECIPROCAL  DIAGRAM 

60.  Framed  Structures.  —  A  framed  structure  is  an  assem- 
blage of  members  for  the  transmission  or  modification  of  external 
forces,  the  internal  stresses  occasioned  thereby  in  the  members 
being  principally  those  of  tension  and  compression.     A  member 
in  tension  is  known  as  a  tie;  if  in  compression,  it  is  known  as  a 
strut. 

A  frame  is  a  theoretical  structure,  the  joints  connecting  its 
members  being  supposed  frictionless.  There  are,  of  course,  no 
such  things  as  frames,  since  all  joints  offer  some  resistance  to 
rotation.  In  general,  however,  engineering  structures  approach 
so  nearly  to  frames  that  no  sensible  error  results  from  treating 
them  as  such. 

The  members  of  a  frame  are  rigid  bars,  hinged  at  the  ends,  and 
it  is  assumed  that:  (a)  The  pins  at  the  joints  are  without  friction. 
(b)  The  external  forces  acting  on  the  frame  are  applied  at  the 
joints. 

If  the  point  of  application  of  a  load  be  at  some  point  inter- 
mediate between  the  joints,  parallel  forces  equivalent  to  the  load 
must  be  substituted  at  the  joints.  If  the  point  of  application 
be  midway  between  joints,  the  equivalent  parallel  forces  at  the 
joints  will  each  be  one-half  the  load;  if  the  point  of  application 
be  otherwise  than  at  the  middle,  then  the  equivalent  parallel 
forces  at  the  joints  may  be  found  by  moments. 

61.  Loads.  —  The  load  on  a  structure  consists  of  the  weight 
of  the  structure  itself  and  the  external  forces  acting  on    it. 
Stationary  and  moving  weights  constitute  dead  and  live  loads 

145 


146          THE   ELEMENTS   OF   MECHANICS  OF   MATERIALS 

respectively.  Wind  pressure,  the  weight  of  the  covering  of  a 
structure  and  the  weights  the  structure  may  support,  are  external 
forces.  The  reactions  of  the  supporting  foundations  of  a  struc- 
ture are  external  forces,  but  they  are  distinguished  from  the 
external  forces  constituting  the  load  by  calling  them  supporting 
forces. 

For  the  equilibrium  of  a  structure  the  external  forces  constitut- 
ing the  load  must  equal  the  supporting  forces,  and  there  must 
be  a  balance  between  the  external  and  the  internal  forces. 

62.  Trusses.  —  Trusses  are  frames  designed  to  support  the 
roofs  of  buildings  and  such  loads  as  are  carried  by  bridges,  the 
supports   being   widely   separated.     There   are   two    classes   of 
trusses:     Those  in  which  the  upper  and  lower  members,  called 
chord  members,  are  parallel  and  horizontal;  and  those  whose 
chords  are  not  parallel.     The  members  connecting  the  upper 
and  lower  chords  are  known  as  braces,  or  as  web  members,  and 
may  be  vertical  or  diagonal.     The  points  at  which  web  members 
meet  a  chord  divide  the  truss  into  bays  or  panels,  and  the  measure- 
ment of  a  bay  is  the  horizontal  distance  between  its  joints. 

The  triangle  being  the  polygon  whose  shape  cannot  be  changed 
without  altering  the  length  of  its  sides,  all  bridge  and  roof  trusses 
are  made  up  of  triangular  frames  as  the  best  means  of  securing 
rigidity. 

Roof  trusses  are  placed  from  10  to  16  feet  apart,  and  each 
truss  is  known  as  a  principal.  The  upper  chord  is  sometimes 
known  as  the  principal  rafter. 

63.  Distinction    between    Beams    and    Girders.  —  Framed 
structures,  such  as  are  used  in  bridges  and  for  the  support  of 
roofs,  are  beams  in  the  sense  that  they  are  supported  at  the  ends 
and  carry  their  loads  between  the  supports,  but  the  term  beam 
seems  to  be  restricted  to  the  cases  where  its  application  is  of 
the  simple  form  and  of  solid  section.     Thus,  the  beam  of  I  sec- 
tion, when  used  alone,  is  known  as  a  beam,  but  when  two  of  them 


FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM         147 

are  compounded  with  plates  riveted  to  their  top  and  bottom 
flanges,  the  combination  is  known  as  a  box  girder,  the  section 
of  which  is  not  solid.  Generally  speaking,  beams  of  built-up 
section  are  girders. 

A  truss  with  its  upper  and  lower  chords  parallel  and  hori- 
zontal is  a  direct  transformation  from  the  simple  I  beam,  the 
chord  members,  like  the  flanges  of  the  7,  resisting  the  bending 
moment;  the  web  members,  like  the  web  of  the  /,  resisting  the 
vertical  shear  and  transmitting  it  from  member  to  member  to 
the  supports. 

If  a  chord  of  a  truss  is  inclined  to  the  horizontal  it  aids  in  the 
transmission  of  the  vertical  shear  and  is  not,  therefore,  designed 
only  to  resist  the  bending  moment. 

64.  Force  Action  at  a  Framed  Joint.  —  A  structure  can  remain 
in  a  state  of  rest  only  when  there  is  equilibrium  in  the  system  of 
forces  acting  on  it,  and  in  order  that  there  shall  be  a  state  of 
rest  at  a  joint  there  must  be  equilibrium  in  the  forces  acting 
on  it  or  transmitted  to  it.  Since  the  joints  are  to  be  considered 
frictionless,  the  external  force  acts  through  the  center  of  the  joint, 
and  the  action  of  a  member  on  the  pin  is  balanced  by  the  reac- 
tion of  the  pin  on  the  member, 
the  action  and  reaction  being 
normal  to  the  surface  of  con- 
tact; and  since  the  joints  are  in 
equilibrium,  the  stresses  in  the 
ends  of  a  member  must  be  equal 
and  opposite,  therefore  the  lines 
of  action  of  the  stresses  in  the 
members  lie  in  the  straight  lines 
joining  the  centers  of  the  pins. 

If  P  be  an  external  force  act- 
ing at  the  joint  A,  Fig.  94,  its  components  along  the  members 
AB  and  AC  (shown  by  the  dotted  lines)  are  equal  and  opposite 


148          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

to  the  stresses  in  the  members.  In  all  cases  the  external  force 
at  a  joint  of  a  frame  must  be  in  equilibrium  with  the  stresses 
in  all  the  members  which  meet  at  the  joint. 

65.  Reciprocal  or  Stress  Diagram.  —  Since  there  is  equilib- 
rium  at  each   joint  of    a  framed   structure,  it    follows  that    a 
closed  polygon  may  be  constructed  whose  sides  represent  the 
acting  forces  in  magnitude  and  direction.     Such  a  polygon  is 
known  as  the  stress  diagram  of  the  frame,  but,  owing  to  its  inter- 
changeable relations  with  the  frame,  it  is  also  known  as  the 
reciprocal  diagram  of  the  frame,  and  is  nothing  more  nor  less 
than  a  polygon  of  forces. 

66.  Frame  Diagram.  —  As  a  preliminary  to  the  construction 
of  the  reciprocal  diagram  a  scale  drawing  of  the  framework  must 
be  made,  which  is  known  as  the  frame  diagram.     The  complete 
preparation  of  the  frame  diagram  comprises  the  following: 

1.  The  determination  of  the  magnitude  and  direction  of  the 
total  load  at  each  joint,  replacing  any  load  that  may  be  applied 
between  two  joints  by  equivalent  parallel  forces  at  the  joints. 
These  equivalent  parallel  forces  are  determined  in  the  same 
manner  as  that  employed  in  determining  the  support  reactions 
of  a  loaded  simple  beam. 

2.  The  determination  of  the  supporting  forces,  or  support 
reactions.     The  reactions  can  be,  and   often   are,  determined 
from  the  funicular  polygon,  but  their  predetermination  affords 
a  check  as  to  the  accuracy,  and  not  infrequently  furnishes  the 
known  external  force  at  a  joint  where  but  two  members  meet, 
thus  providing   a   starting  point  for   the   construction   of   the 
reciprocal  diagram. 

3.  The  correct  lettering  of  the  diagram  according  to  the  Bow 
system. 

4.  The  marking  with  arrowheads  of  all  the  external  forces, 
giving  to  each  its  value,  and  taking  care  that  the  arrowheads  do 
not  cross  the  lines  of  the  frame  diagram. 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM 


149 


67.  Support  Reactions.  —  With  a  small  span  the  ends  of  a 
roof  truss  are  fixed,  and  the  reactions  at  the  supports  are  vertical 
when  the  loads  are  vertical.  In  case  of  a  wind  load,  acting  only 
on  one  side  of  the  roof,  the  reactions  due  to  it  act  in  directions 
parallel  to  the  normal  wind  pressure. 

With  large  spans  one  end  of  the  truss  is  fixed  while  the  other 
end  is  on  rollers,  thus  permitting  a  lateral  movement  in  case  of 
expansion.  The  reaction  at  the  fixed  end  will  be  inclined  and 
that  at  the  free  end  will  be  vertical. 

It  should  be  noted  that  when  all  the  external  forces  are  ver- 
tical the  support  reactions  are  vertical.  In  such  cases  the  loads 
that  come  directly  over  the  supports  are  omitted,  as  they  have 
no  influence  on  the  stresses  in  the  members.  Such  loads  must 
be  deducted  from  the  total  support  reactions  in  order  to  obtain 
the  proper  reactions  to  use  in  the  determination  of  the  stresses. 


1+2= 3  tons 


Iton 


2  tons 


R2=5  tons 


Fig-  95- 


For  example,  Fig.  95  represents  a  king-post  truss  with  a  uni- 
form load  of  8  tons  on  one  side  and  4  tons  on  the  other.  The 
loads  are  applied  to  the  joints  as  follows: 

Of  the  4  tons  on  the  left-hand  rafter  we  may  assume  one-half 
of  it  borne  by  the  member  AF  and  the  other  half  by  the  member 
EG.  Distributing  these  two  loads,  giving  half  of  each  to  the 
ends  of  the  member  which  supports  it,  we  get  a  total  load  of 


THE   ELEMENTS   OF   MECHANICS   OF   MATERIALS 


2  tons  at  joint  i,  i  ton  at  the  apex  end  of  the  rafter,  and  i 
ton  over  the  left  support.  Proceeding  in  a  similar  manner  with 
the  right-hand  rafter,  we  get  a  total  of  3  tons  at  the  apex  of  the 
rafters,  4  tons  at  the  joint  3,  and  2  tons  over  the  right  support. 
To  find  the  support  reactions  we  reject  the  loads  over  the 
supports,  and,  assuming  the  span  to  be  a,  take  moments  thus : 


X  a  =  2  X 


R2Xa  =  4X  — 
4 


d  d  -„ 

3  X  -  +  4  X  -  ,  whence  RI  =  4  tons. 

3  X  — h  2  X  -  ,  whence  R2  =  5  tons. 
2  4 


These  values  of  RI  and  R2  are  to  be  used  in  determining  the 
stresses  in  the  members,  but  the  total  wall  reactions  are  5  tons 
and  7  tons  for  RI  and  R2  respectively. 

68.  Wind  Pressure.  —  For  purposes  of  computation  the  direc- 
tion of  the  wind  is  assumed  to  be  horizontal,  and  its  intensity 
may  be  taken  as  40  pounds  per  square  foot. 

Roofs  generally  present  an  inclined  surface  to  the  wind,  but 
as  the  roof  structure  itself  must  resist  the  normal  pressure  to 
which  it  is  subjected,  it  is  important  to  know  this  normal  pres- 
sure for  the  different  degrees  of  roof  inclination. 

The  normal  pressures  due  to  a  horizontal  intensity  of  40  pounds 
per  square  foot  on  a  vertical  surface  have  been  determined  experi- 
mentally for  roofs  of  different  inclinations,  and  are  set  forth  in 
the  table  which  follows: 


Pitch  of  roof  in 
degrees. 

Normal  pressure  in  pounds 
per  square  foot. 

5 

5 

10 

10 

15 

U 

20 

18 

25 

22 

30 

26 

35 

30 

40 

33 

45 

36 

50 

38 

55 

39 

60 

40 

FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM          151 

For  horizontal  pressures  other  than  40  pounds  the  normal 
pressures  are  directly  proportional  to  those  given  in  the 
table. 

69.  Drawing  the  Reciprocal  Diagram.  —  In  drawing  the  recip- 
rocal diagram  we  proceed  as  follows: 

1 .  Select  a  scale  —  as  large  as  practicable  —  and  draw  the 
force  diagram  of  the  external  forces,  marking  the  beginning  and 
ending  of  a  line  representing  a  force  with  the  small  letters  of  the 
alphabet  corresponding  to  the  capital  letters,  taken  in  clockwise 
order,  found  in  the  spaces  flanking  the  force  in  the  frame  diagram. 
This  diagram  represents  the  magnitudes  and  directions  of  the 
external  loads  on  the  joints  and  of  the  support  reactions,  and, 
if  properly  drawn,  forms  a  closed  polygon.     In  the  most  frequent 
case,  that  of  vertical  loads,  the  polygon  resolves  itself  into  a 
straight  line,  vertical  in  direction,  as  already  explained.     Deter- 
mine the  magnitude  of  the  support  reactions,  either  by  moments 
or  by  the  method  of  the  funicular  polygon. 

2.  Choose  as  a  starting  point  a  joint  where  a  sufficient  number 
of  conditions  are  known  to  enable  its  reciprocal  to  be  drawn, 
being  careful  to  letter  the  lines  of  the  reciprocal  with  the  small 
letters   of    the   alphabet    corresponding   to    the   capital   letters 
denoting  the  members  in  the  frame  diagram.     The  lengths  of 
the  lines  of  this  reciprocal,  to  the  chosen  scale,  give  the  magni- 
tudes of  the  stresses  in  the  members  to  which  they  refer,  and  the 
directions,  or  the  kinds,  of  these  stresses  are  at  once  determined 
by  noting  the  directions  in  which  the  successive  lines  of  the 
reciprocal   were   drawn.     The   directions   of   the   stresses   thus 
found  are  at  once  indicated  by  placing  arrowheads  on  the  mem- 
bers in  the  frame  diagram.     Since  there  are  equal  and  opposite 
stresses  in  the  ends  of  a  member,  arrowheads  must  now  be  placed 
on  the  other  ends  of  the  members  whose  stresses  have  been 
found,  making  them  point  in  the  opposite  direction  to  those 
placed  at  the  joint  whose  reciprocal  has  been  drawn. 


152 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


3.  Proceed  by  similar  processes  until  the  reciprocals  of  all 
the  joints  are  drawn  and  the  stresses  in  all  the  members 
determined. 

It  should  be  noted  that  the  reciprocal  of  a  joint  cannot  be 
drawn  if  the  stresses  in  more  than  two  of  the  members  forming 
it  are  unknown.  In  general  terms,  the  drawing  of  the  reciprocal 
of  a  joint  of  n  forces  involves  2  n  conditions,  viz.7  n  magnitudes 
and  n  directions,  and  unless  2  n  —  2  of  these  conditions  are 
known  the  data  is  insufficient  to  draw  the  reciprocal. 


3  tons 


j=3  tons 


3  tons 


R2=3  tons 


Fig.  96. 


Example  I.  —  Suppose  each  rafter  of  the  roof  truss  of  Fig.  96 
to  support  an  evenly  distributed  load  of  6  tons;  it  is  required 
to  find  the  magnitudes  and  kinds  of  stresses  in  the  members  of 
the  truss. 

We  commence  by  apportioning  the  loads  to  the  joints  of  the 
frame  diagram.  In  doing  so,  we  find  a  total  load  of  6  tons  at 
the  ridge  and  a  load  of  3  tons  directly  over  each  supporting  wall. 
As  these  latter  do  not  affect  the  stresses  in  the  members  they 
will  be  omitted  from  any  further  discussion,  remembering  that 
if  the  total  reactions  at  the  walls  are  required,  RI  and  R2  must 
each  be  increased  by  3  tons. 


FRAMED   STRUCTURES  —  RICIPROCAL  DIAGRAM  153 

The  truss  being  symmetrical,  the  support  reactions  RI  and  RI 
are  each  equal  to  half  of  the  load  of  6  tons  at  the  ridge. 

Letter  the  frame  diagram  as  shown,  ignoring  the  vertical  loads 
over  the  supports,  so  that  R2  will  be  known  as  BC  and  RI  as  CA . 
Adopt  a  load  scale  of  0.25  inch  to  the  ton. 

Since  each  of  the  three  joints  of  the  frame  has  but  two  members 
and  a  known  external  force,  either  may  be  used  as  a  starting 
point.  Selecting  the  joint  at  the  left  support,  we  draw  ca  equal 
in  length  to  0.75  inch  to  represent  the  left  reaction  CA  of  3  tons, 
and  we  measure  it  upward  from  c  to  a  because  RI  acts  upward. 
From  a  and  c  draw  lines  parallel  to  the  members  AD  and  DC 
respectively.  They  intersect  at  J,  giving  the  triangle  cad  as 
the  triangle  of  forces,  or  the  reciprocal  diagram,  for  the  joint  at 
the  left  support.  Hence,  ad  and  dc  represent  in  magnitude  and 
direction  the  stresses  in  the  members  AD  and  DC.  Knowing 
the  direction  of  RI,  as  represented  by  ca,  the  directions  of  the 
actions  of  the  stresses  in  the  two  members  are  known  by  taking 
the  sides  of  the  triangle  in  order,  as  shown  by  the  arrows.  The 
stress  in  AD  acts  in  the  direction  ad,  and  that  in  DC  in  the 
direction  dc.  Indicate  these  directions  by  placing  arrowheads 
on  the  two  members  at  points  close  to  the  joint.  Knowing  that 
the  stress  in  one  end  of  a  member  is  opposed  by  an  equal  and 
opposite  stress  in  the  other  end,  we  may  now  place  arrowheads 
at  the  other  ends  of  the  members  AD  and  DC  accordingly,  as 
shown. 

To  draw  the  reciprocal  of  the  joint  at  the  right  support  we 
draw  be  upward,  and  make  it  0.75  inch  in  length  to  represent 
the  3  tons  of  the  right  reaction  BC  in  magnitude  and  direction. 
From  c  and  b  draw  lines  parallel  to  the  members  CD  and  DB 
respectively.  They  intersect  at  d,  giving  the  triangle  bed  as 
the  reciprocal  of  the  joint  at  the  right  support.  Hence,  cd  and 
db  represent  in  magnitude  and  direction  the  stresses  in  the  mem- 
bers CD  and  DB.  The  arrows  within  the  triangle  bed  show  the 


THE   ELEMENTS   OF  MECHANICS  OF   MATERIALS 


direction  of  these  stresses,  and  they  are  marked  with  arrowheads 
accordingly  in  the  frame  diagram.  By  scale  measurements  of 
the  reciprocals  the  stresses  in  the  rafters  are  found  to  be  4.4  tons 
each,  and  in  the  tie  CD  the  stress  is  found  to  be  3.2  tons. 

If  to  the  frame  of  Fig.  96  we  add  a  vertical  rod,  called  the  king- 
post, we  obtain  the  frame  of  Fig.  97. 


6  tons 


!=  3  tons 


Ro=3  tons 


Fig.  97. 


To  draw  the  reciprocal  of  the  joint  ABED  at  the  ridge  we 
measure  ab  downward,  and  make  it  1.5  inches  long  to  represent, 
to  the  chosen  scale  of  0.25  inch  to  the  ton,  the  magnitude  and 
direction  of  the  external  force  of  6  tons.  The  stress  in  the 
rafter  AD  has  been  found  to  be  4.4  tons,  so  we  draw  ad  parallel 
to  AD  and  make  it  i.i  inches  long  to  represent  the  4.4  tons  to 
scale.  From  b  we  draw  a  line  parallel  to  BE  and  we  find  it 
intersects  ad  at  d.  This  shows  that  d  and  e  are  one  and  the 
same  point,  and  that  there  is  no  stress  in  the  vertical  post  DE. 
The  purpose  of  DE  is  to  prevent  sagging  in  the  horizontal 
member. 

As  a  further  illustration  we  will  consider  the  king-post  truss 
of  Fig.  95,  reproduced  in  Fig.  98.  The  loads  and  support  reac- 
tions were  found  to  be  as  shown. 

Knowing  the  force  EA,  or  RI,  we  commence  at  the  joint  at 


FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM          155 


Dl  =10.0 
IH=4.0. 


Fig.  99. 


156          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

the  left  support  and,  with  a  scale  0.25  inch  to  the  ton,  obtain  eaf 
as  the  triangle  of  forces  for  the  joint.  Marking  with  arrowheads 
the  kinds  of  stresses  in  the  members  AF  and  FE,  we  proceed  to 
the  next  joint  in  clockwise  order,  that  at  which  the  load  of  2  tons 
is  applied. 

Of  the  eight  conditions  of  this  joint  the  four  directions  and 
two  of  the  magnitudes  are  known,  the  magnitude  of  FA  having 
just  been  determined.  We  then  readily  obtain  abgf  as  the  force 
polygon  of  the  joint.  Marking  with  arrowheads  the  kinds  of 
stresses  in  the  members  BG  and  GF,  we  proceed  in  clockwise 
order  to  the  other  joints,  obtaining  for  the  joint  at  the  ridge  the 
force  polygon  gbch;  for  the  joint  HCDI  the  polygon  kcdi\  and 
for  the  joint  at  the  right  support  the  triangle  ide,  thus  completing 
the  determination  of  the  stresses  in  all  the  members. 

It  will  be  observed  that  each  of  the  force  polygons  of  the 
joints,  when  taken  in  order,  contains  a  side  of  the  one  immediately 
preceding  it;  hence,  each  polygon  can  be  built  upon  the  one  pre- 
ceding it,  and  so  produce  one  figure  which  will  contain  all  the 
sides  and  be  a  graphic  representation  of  the  magnitudes  and  direc- 
tions of  all  the  external  forces  and  internal  stresses  of  the  struc- 
ture. Such  a  figure  is  the  reciprocal  diagram  of  the  structure. 

The  reciprocal  diagram  of  the  truss  of  Fig.  98  is  shown  in 
Fig.  99,  the  load  line  ad  having  first  been  drawn  and  the  lines 
representing  the  stresses  then  drawn  in  their  regular  order. 
The  magnitudes  of  the  stresses  were  measured  to  the  scale  and 
found  to  be  as  shown  in  the  table. 

Any  attempt  to  put  arrowheads  on  the  reciprocal  diagram  to 
indicate  the  kinds  of  stresses  in  the  members  results  in  nothing 
but  confusion.  If  the  kind  of  stress  in  a  member  cannot  be  dis- 
covered by  the  eye,  the  polygon  of  forces  for  the  joint  in  question 
must  be  drawn,  as  was  done  in  connection  with  Fig.  98. 

70.  Rule  for  Determining  the  Kind  of  Stress  in  a  Member. 
• —  If,  after  determining  the  directions  of  the  stresses  in  all  the 


FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM          157 

members  by  means  of  the  reciprocals  of  the  joints,  and  after 
marking  the  ends  of  the  members  with  arrowheads  accordingly, 
it  is  found  that  the  arrowheads  of  a  member  point  toward  its 
joints,  the  member  is  then  in  compression  and  is  a  strut ;  if  they 
point  away  from  the  joint  the  member  is  in  tension  and  is  a  tie. 
Thus  it  is  found  that  the  rafters  AD  and  BD,  Fig.  96,  are  in 
compression,  and  the  member  CD  is  in  tension. 

71.  Method  of  Sections  in  Determining  Stresses.  —  This 
method  depends  upon  the  principle  demonstrated  in  Art.  58, 
that  at  any  imaginary  section  of  a  frame  there  is  equilibrium 
between  the  external  forces  on  one  side  of  the  section  and  the 
stress  forces  in  members  on  the  same  side  that  are  cut  by  the 
section,  and  that,  therefore,  the  algebraic  sum  of  the  moments 
about  any  point  in  the  plane  of  the  frame  must  be  zero.  There 
must,  of  course,  be  but  one  unknown  force  in  the  equation  of 
moments,  and  this  will  be  the  case : 

(a)  When  only  two  members  are  cut  and  one  of  them  passes 
through  the  center  of  moments. 

(b)  When  three  members  are  cut  and  the  center  of  moments  is 
taken  at  the  intersection  of  two  of  them. 

(c)  When  the  stresses  are  known  in  all  the  members  cut 
except  one. 

For  example,  the  roof  truss  of  Fig.  98  is  reproduced  in  Fig.  100. 
Knowing  the  magnitude  and  direction  of  the  left  reaction  RI, 
and  considering  the  equilibrium  of  the  joint  at  the  left  support, 
it  is  seen  from  inspection  that  AF  is  in  compression  and  FE 
in  tension,  and  they  are  so  marked  with  arrowheads  in  the 
frame. 

There  being  equilibrium  at  the  joint,  we  have  the  static  equa- 
tions, Si  cos  30°  =  5*2  and  Si  sin  30°  =  RI,  in  which  Si  and  S% 
are  the  stresses  in  AF  and  FE  respectively.  From  these  equa- 
tions we  get 

Si  =  stress  in  AF  =  8  tons,  and  52  =  stress  in  FE  =  6.9  tons. 


158 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


Knowing  the  stress  in  AF  and  the  external  force  AB,  an  inspec- 
tion of  joint  ABGF  shows  BG  and  GF  to  be  struts,  and  they  are 
so  marked  in  the  frame.  The  section  xy  cuts  the  members  BG, 
GF,  and  FE,  two  of  these  members  intersecting  at  E.  Consider- 
ing the  part  of  the  truss  to  the  left  of  xy,  and  denoting  the  stress 
in  BG  by  53,  we  have,  by  moments  about  the  joint  at  the  middle 
of  the  lower  chord, 
53  X  12.5  sin  30°  +  2  X  6.25  =  4  X  12.5,  whence  6*3  =  6  tons. 


3  tons 


R2  =  5  tons 


S:  cos  30 


Fig.  100. 

Calling  5*4  the  stress  in  GF,  and  taking  moments  about  the  left 
support,  we  have 

S4  X  12.5  sin  30°  =  2  X  6.25,   whence   ^4  =  2  tons. 

These  results  are  the  same  as  were  obtained  from  the  reciprocal 
diagram  of  Fig.  99. 

In  the  examples  of  the  section  method  just  given  the  nature 
of  the  stress,  whether  tension  or  compression,  in  the  member 
whose  stress  was  sought  was  known,  so  that  the  nature  of  the 
moments,  whether  clockwise  or  contraclockwise,  about  the 
center  of  moments  was  known.  It  is  not,  however,  essential 
that  the  nature  of  the  stress  in  a  member  be  known  in  order  to 


FRAMED   STRUCTURES  —  RECIPROCAL   DIAGRAM          159 

determine  it.  A  member  may  be  assumed  to  be  either  in  ten- 
sion or  in  compression  and  the  equation  of  moments  written 
accordingly.  Should  the  solution  of  the  equation  give  a  positive 
result  for  the  stress,  the  assumption  of  its  nature  is  the  correct 
one;  should  the  resulting  stress  be  negative  in  sign,  then  the 
assumption  as  to  the  nature  of  the  stress  is  incorrect,  but  the 
numerical  value  of  the  stress  will  be  correct  and  the  same  as  in 
the  first  instance.  These  conditions  arise  from  the  fact  that, 
in  the  first  instance,  the  moment  of  the  required  stress  was 
placed  in  the  proper  member  of  the  equation  of  moments;  in 
the  second  instance  it  was  improperly  placed. 

For  example,  in  finding  the  stress  in  BG  we  will  assume  it  to 
be  one  of  tension.  The  equation  of  moments  about  the  middle 
of  the  lower  chord  as  a  center  will  be 

4  X  12.5  -fS3  X  12. 5  sin 30°=  2  X  6.25,  whence  S3=  —  6  tons, 
a  negative  result,  which  shows  that  BG  is  in  compression  and  not 
in  tension,  as  was  assumed,  but  the  numerical  result  is  the  same 
as  found  above. 

Again,  assuming  FE  to  be  in  tension,  we  shall  have,  with 
moments  about  the  joint  at  the  middle  of  the  left  rafter, 

Sz  X  6.25  tan  30°  =  4  X  6.25,  whence  62  =  6.9  tons, 

as  was  found  above,  the  positive  result  showing  that  FE  was 
correctly  assumed  to  be  in  tension. 

72.  Stresses  in  Braced  Cantilevers.  —  The  stresses  in  the 
members  of  the  braced  cantilever  of  Fig.  101,  having  a  con- 
centrated load  of  2  tons  at  its  outer  extremity,  may  be  found 
as  follows: 

To  the  scale  of  5V  inch  =  iV  ton  lay  off  ab  vertical  and  i  inch 
in  length  to  represent  the  load  of  2  tons.  From  b  draw  a  parallel 
to  EC  and  from  a  a  parallel  to  CA .  They  intersect  at  c.  Then, 
abc  is  the  triangle  of  forces  for  the  equilibrium  of  the  joint  ABC. 
The  force  AB,  acting  downward,  is  denoted  in  direction  and 


i6o 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


magnitude  by  ab  in  the  triangle,  and  the  stresses  in  the  other 
members  taken  in  order  act  in  the  directions  be  and  ca,  and  their 
magnitudes  are,  of  course,  denoted  by  the  lengths  of  these  lines 
to  the  chosen  scale. 


2  tons 


Mark  these  directions  by  arrowheads  at  the  joint  ABC.  The 
directions  of  the  stress  in  the  upper  end  of  BC  and  in  the  outer 
end  of  CA  being  now  known,  it  follows  that  the  stresses  in  the 
other  ends  of  these  members  are  equal  and  opposite,  and  arrow- 
heads must  be  at  once  placed  to  indicate  them.  We  now  have 
sufficient  data  to  draw  the  force  diagram  of  the  joint  CBD.  The 
stress  in  the  lower  end  of  CB  has  just  been  marked  to  act  in  the 
direction  cb.  From  b  and  c  draw  parallels  to  BD  and  DC  re- 
spectively, intersecting  at  d\  then  the  triangle  cbd  is  the  force 
diagram  for  the  joint  CBD,  and  bd  and  dc  are  the  directions  of 
the  stresses  in  the  members  BD  and  DC,  and  they  have  been 
marked  accordingly  in  the  frame  diagram.  Arrowheads  are  at 
once  placed  at  the  other  ends  of  BD  and  DC  to  denote  the 
directions  of  the  equal  and  opposite  stresses  at  the  wall  and  at 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM          161 

the  joint  ACDE  respectively.  The  stresses  in  AC  and  CD  at 
the  joint  ACDE  are  now  known  and  their  directions  are  ac  and 
cd  respectively.  From  d  draw  a  parallel  to  DE  and  from  a 
a  parallel  to  EA.  They  intersect  at  et  and  de  and  ea  are  the 
directions  and  magnitudes  of  the  stresses  in  the  members  DE 
and  EA  respectively.  It  should  be  noted  that,  in  the  reciprocal 
diagram,  the  stress  line  of  one  member  may  lie  wholly  or  partly 
on  the  stress  line  of  another  member,  as  was  here  instanced  in 
the  stresses  of  AC  and  EA.  The  stresses  in  the  different  mem- 
bers were  measured  to  scale  on  the  reciprocal  diagram  and  marked 
on  the  members  of  the  frame. 

The  horizontal  outward  pull  of  2.95  tons  at  the  upper  joint  at 
the  wall  occasions  an  equal  and  opposite  reaction.  At  the  lower 
wall  joint  there  is  a  horizontal  thrust  of  2  tons  and  a  diagonal 
thrust  of  2.25  tons.  The  horizontal  component  of  this  diagonal 
thrust  is  hd  =  ac  =  0.95  ton,  making  the  two  reactions  equal 
but  opposite  in  direction. 

The  braced  cantilever  of  Fig.  102  is  25  feet  long,  10  feet  deep 
and  uniformly  loaded  on  the  top  with  100  pounds  per  foot  run. 

In  apportioning  the  total  load  of  2500  pounds,  it  will  be 
observed  that  the  member  BJ,  being  but  half  the  length  of  each 
of  the  members  CH  and  DF,  sustains  but  one-fifth  of  the  total 
load,  CH  and  DF  each  sustaining  two-fifths. 

The  apportionment  of  the  load  will  therefore  be:  500  pounds 
at  the  joint  at  the  outer  end,  1000  pounds  at  the  joint  CDFGH, 
750  pounds  at  the  joint  BCHIJ ',  and  250  pounds  at  the  joint 
ABJ  at  the  wall.  This  latter  load  has  no  influence  on  the 
stresses  in  the  members  and  is  rejected. 

Commencing  at  the  joint  DEF,  and  with  a  scale  of  ^V  inch 
=  100  pounds,  we  get  def  as  the  force  diagram  for  the  equilibrium 
of  the  joint,  and  at  once  mark  arrowheads  on  the  frame  indicating 
the  directions  of  the  stresses  EF,  FD,  FE,  and  DF.  Having  now 
the  magnitude  of  the  stress  FE  at  the  joint  FEG,  we  obtain 


162 


THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 


feg  as  the  force  diagram  of  the  joint.  Marking  with  arrowheads 
on  the  frame  the  directions  of  the  stresses  thus  obtained,  we  find 
that  at  the  joint  CDFGH  the  force  CD  and  the  stresses  in  DF 
and  FG  are  known,  and  therefore  the  force  polygon  cdfgh  is 
readily  obtained.  The  reciprocal  diagram  may  now  be  com- 
pleted without  difficulty. 


Fig.  102. 

73.  The  Warren  girder  of  Fig.  103  has  a  span  of  27  feet,  and 
is  loaded  with  ij  tons  per  foot,  making  a  total  load  of  36  tons. 
The  members  AF  and  DL  each  sustains  but  one-sixth  of  the 
load,  and,  in  the  apportionment,  one-twelfth,  or  3  tons,  falls 
over  the  support  at  each  end,  and  the  remainder  of  the  load  as 
shown. 

At  each  end  of  the  top  flange  there  is  equilibrium  under  the 
action  of  the  external  force  of  3  tons  in  line  with  the  upright 
member  and  the  stresses  in  the  two  members  at  right  angles  to 
each  other.  Evidently  the  stress  in  each  of  the  uprights  is 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM 


163 


3  tons,  therefore  there  cannot  be  any  stress  in  the  members  AF 
and  DL  if  the  equilibrium  is  to  be  maintained.  To  indicate 
that  there  is  no  stress  in  AF  the  letters  a  and/  must  be  placed 
at  the  same  point  in  the  reciprocal  diagram,  and  because  of  the 
absence  of  stress  in  DL  the  letters  d  and  /  are  at  the  same  point. 
It  should  not  be  forgotten  that  the  two  external  forces  of  3  tons 
at  the  ends,  coming  directly  over  the  supports,  are  neglected,  and 


3  tons       9  itons 

A      J  B 


ISitons 

Y 


9, tons       3  tons 
f    D 


15  tons 


STRESSES 
FG  =  KL= 17.32  tons 
GE=  KE  =  8.66  tons 
GH  =  H I  =  I J  =J  K=7.5"  tons 
El— 16.2  tons 
BH  =  CJ =12.5  tons 


15  tons 


- — ->-S  sin  30 


Fig.  103. 

that  the  reactions  of  15  tons  each  are  due  solely  to  the  loads  of 
9  tons,  1 2  tons,  and  9  tons,  the  forces  which  occasion  the  stresses 
in  the  members.  In  the  construction  of  the  Warren  girder  the 
end  uprights  and  the  members  AF  and  DL  are  omitted. 

To  the  scale  of  J  inch  =  3  tons  set  off  ef  to  represent  in  mag- 
nitude and  direction  the  left  reaction  EF  of  15  tons.  From/ 
and  e  draw  parallels  to  FG  and  GE  respectively,  intersecting  at  g. 
Then,  efg  is  the  triangle  of  forces  for  the  joint  at  the  left  support. 


164 


THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 


For  the  joint  GFABH  we  get  gfabh  as  the  polygon  of  forces,  and 
so  on  to  the  completion  of  the  reciprocal  diagram. 

The  triangles  of  the  Warren  girder  being  equilateral,  the  stresses 
found  from  the  reciprocal  diagram  may  easily  be  checked.  Thus, 
denoting  the  stress  in  LK  by  5,  we  have 

S  =  15  sec  30°  =  17.32  tons. 

74.  The  Linville  or  N  girder  of  Fig.  104  is  irregularly  loaded 
on  the  bottom  flange  as  shown. 


cl 


STRESSES 

CL=IH=EA=O;    LD=32.5tons; 


FO  =  G  A  =  FG  =  D  A  =  17.5  tons. 
KL  -=  46  tons  ;   K  J  =  2.5  tons; 
DK  =  JB  =31.5  tons; 

Jl  =  4tons;    DI  =  DH=  34.5  tons. 
Fig.   104. 


For  convenience  the  frame  is  lettered  in  this  instance  in  con- 
traclockwise  order  and  the  forces  at  the  joints  will  be  considered 
in  like  manner. 

To  a  scale  of  ?V  inch  =  i  ton,  set  off  ab  and  be  to  represent 
in  magnitude  and  direction  the  loads  of  20  tons  and  30  tons 
respectively. 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM          165 

Selecting  a  pole  o  we  obtain,  by  means  of  the  funicular  polygon, 
cd  and  da  for  the  reactions  R%  and  RI  respectively. 

For  the  reason  already  given  there  can  be  no  stress  in  the 
members  CL  and  EA,  and  this  is  indicated  by  placing  I  in  the 
reciprocal  diagram  at  the  same  point  as  c,  and  e  at  the  same 
point  as  a. 

At  the  joint  at  the  middle  of  the  top  flange  there  is  a  state  of 
equilibrium  under  the  action  of  the  stresses  in  the  members  ID, 
DH,  and  HI,  HI  being  at  right  angles  to  each  of  the  others. 
There  cannot,  therefore,  be  any  stress  in  HI,  and  h  and  i  will 
fall  at  the  same  point  in  the  reciprocal  diagram. 

The  members  CL  and  EA  add  rigidity  to  the  frame,  and  HI 
resists  the  tendency  of  the  top  flange  to  bend. 

For  the  equilibrium  of  the  joint  LDK  we  get  the  triangle  Idk. 
For  the  joint  CLKJB  we  get  the  force  polygon  clkjb,  and  so  on 
to  the  completion  of  the  reciprocal  diagram. 

75.  The  Fink  Truss.  —  The  Fink  truss  of  four  bays,  Fig.  105, 
has  a  span  of  64  feet,  a  depth  of  12  feet,  and  is  uniformly  loaded 
with  1.5  tons  per  foot,  making  96  tons  in  all.  The  figure  is 
constructed  to  a  scale  of  -^  inch  to  the  foot,  and  the  load  scale 
is  taken  as  ^  inch  to  the  ton.  The  apportionment  of  the  load 
places  12  tons  over  each  support  and  they  are  rejected. 

In  the  construction  of  the  reciprocal  diagram  there  are  in- 
sufficient data  to  begin  at  either  of  the  support  joints.  A  con- 
sideration of  the  equilibrium  of  joint  2  shows  the  stress  in  FI  to 
be  24  tons.  At  joint  n  we  have  four  forces  acting  along  two 
lines;  therefore  the  conditions  of  equilibrium  require  that  the 
stress  in  FI  shall  equal  that  in  HG,  and  that  the  stress  in  GF 
shall  equal  that  in  IH.  Then  the  stress  in  HG  is  also  24  tons. 
At  joint  10  we  have  equilibrium  under  the  action  of  the  stresses 
in  GH,  HE,  and  EG]  and  since  HE  and  EG  are  equally  inclined 
toGH  their  stresses  must  be  equal.  If,  then,  the  stress  in  HE 
can  be  found,  the  stress  in  EG  will  be  known,  and  there  will  be 


1 66 


THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 


sufficient  data  to  begin  at  the  joint  at  the  left  support.  Since 
GH  is  a  strut  it  is  evident  that  the  equilibrium  of  joint  10  re- 
quires that  HE  and  EG  be  ties,  and  therefore  the  direction  of 
the  stress  in  HE  at  joint  10,  and  also  the  direction  of  the  stress 
in  EG  at  joint  i,  are  known.  By  means  of  the  section  xy  and 
moments  about  the  left  support  we  have 

Stress  in  HE  X  32  sin  9  =  24  X  16, 
whence,        Stress  in  HE  =  stress  in  EG  =  20  tons. 


24tons 


24  tons 


24!  tons 


16' 


36  tons  \ 


There  are  now  sufficient  data  to  begin  the  construction  of  the 
reciprocal  diagram  by  a  consideration  of  the  equilibrium  of  the 
joint  at  the  left  support. 

Set  off  the  load  line  ad,  making  ab,  be,  and  cd  each  equal  to 
ff  inch  in  length  to  represent  the  external  loads  of  24  tons. 
Then  de  and  ea  are  the  right  and  left  reactions  respectively,  to 
scale. 


FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM          167 

From  e  draw  eg  parallel  to  EG,  and  make  it  f  %  inch  long  to 
represent  the  stress  of  20  tons  in  EG.  Commencing  at  e,  we  get 
eafg  as  the  stress  polygon  for  the  joint  at  the  left  support.  Know- 
ing of,  the  stress  polygon  abifior  the  joint  ABIF  is  readily  drawn, 
and  so  on  to  the  completion  of  the  reciprocal  diagram. 

By  means  of  the  section  ut  and  moments  about  the  upper 
middle  joint,  we  obtain 

Stress  in  JE  X  32  sin  a  +  24  X  16  =  36  X  32, 
whence 

Stress  in  JE  =  (36X32-24X16)^  _  68.3S  tons> 

which  agrees  with  the  scale  length  oije  of  the  reciprocal  diagram. 
The  stresses  in  AF,  BI,  CL,  and  DO  are  shown  by  the  diagram 
to  be  80  tons  each.     This  may  be  checked  by  the  sections  ut 
and  vz. 

Stress  in  BI  X  12  =  36  X  16  +  stress  in  JE  X  6  cos  a, 
whence 

Stress  in  BI  =  48  +  68.35  X  -4=.  =  80  tons. 

^73 

From  the  section  vz  and  moments  about  the  joint  FIHG  we 
have 
Stress  in  EG  X  2  A/73  sin  (d  -  a)  +  stress  in  AF  X  6  =36  X  16. 

whence,  Stress  in  4F  =  57    ~  9    =  80  tons. 

6 

It  will  be  noted  that  the  Fink  truss  is  composed  of  a  primary 
truss  i,  8,  5,  and  two  secondary  trusses  i,  10,  3,  and  3,  6,  5. 
Whether  or  not  there  are  joints  at  u,  9,  7,  and  12  will  not  affect 
the  stresses  in  GF,  FI,  IE,  EG,  IJ,  JE,  KL,  LM,  ME,  EK, 
LO,  ON,  NM. 

76.   Roof  Truss  Fixed  at  the  Ends  and  with  Wind  Pressure.  - 
The  roof  truss  of  Fig.  106  is  fixed  at  the  ends  and  has  the  follow- 
ing data: 


1 68          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 


Fig.  106. 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM          169 

Linear  scale,  i  inch  =  10  feet;  load  scale,  0.75  inch  =  i  ton. 
Pitch  of  roof,  30°.  Span  of  roof,  30  feet.  Distance  between 
trusses,  10  feet.  Dead  load  per  square  foot  of  horizontal  sur- 
face, 23  pounds.  Horizontal  wind  pressure  per  square  foot, 
40  pounds. 

The  dead  vertical  load  =  3- 3  _  ?  tons,  approximately, 

2240 

and  its  uniform  distribution  places  0.75  ton  at  each  of  the 
joints  2,  3,  and  4,  and  0.375  ton  at  each  of  the  joints  i  and  5. 

In  this  case  of  a  truss  fixed  at  the  ends,  the  effect  of  the  wind 
and  of  the  vertical  loads  at  the  end  joints  is  borne  entirely  by 
the  supports  and  does  not  affect  the  stresses  in  the  members; 
the  loads  at  the  supports  are  therefore  omitted  from  consider- 
ation, and  the  reactions  to  be  used  are  those  due  only  to  the  loads 
producing  the  stresses. 

The  effective  support  reactions  due  to  the  vertical  loads  are 

each —  =  1.125    tons  =  1.125  X  0.75  =  f£   inch    when 

2 

reduced  to  scale.  These  reactions  are  vertical  and  are  denoted 
in  the  frame  diagram  by  Rv. 

The  vertical  loads  at  the  joints  2,3,  and  4  are,  when  reduced 
to  scale,  each  equal  to  0.75  X  0.75  =  T9e  inch.  These  vertical 
loads  are  laid  off  to  scale  at  the  joints. 

Suppose  the  wind  to  act  on  the  left  side  of  the  roof.  The 
pitch  of  the  roof  being  30°,  and  the  horizontal  intensity  of 
the  wind  pressure  being  40  pounds  per  square  foot,  the  normal 
pressure  is  26  pounds  per  square  foot  (see  table  of  Art.  68). 

The  total  wind  pressure  to  be  borne  by  one  truss  is 

i«;  sec  30°  X  10  X  26  , 

-* =  2  tons,  very  nearly. 

2240 

The  distribution  of  this  wind  load  places  i  ton  at  joint  2  and 
I  ton  at  joint  3,  the  half  ton  at  the  left  support  being  rejected. 
These  loads  to  scale  are  J  inch  and  f  inch  respectively,  and  are 


1 70          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

combined  with  the  vertical  loads  so  as  to  obtain  the  magnitudes 
and  directions  of  the  resultant  pressures  at  the  joints. 

The  truss  being  fixed  at  the  ends  the  support  reactions  due  to 
the  wind  pressure  must  be  parallel  to  and  opposite  in  direction 
to  the  normal  wind  pressure  in  order  that  there  shall  be  a  balance 
between  the  external  forces.  The  magnitude  of  these  effective 
wind  reactions  can  be  found  by  means  of  a  funicular  polygon. 
Thus: 

Lay  off  the  load  line  a!c'  parallel  to  the  normal  wind  pressure, 
making  a'bf  equal  to  the  wind  pressure  at  joint  2,  and  b'cf  equal 
to  the  wind  pressure  at  joint  3.  Select  at  random  a  pole  0,  and 
draw  the  vectors  oaf,  ob' ',  and  ocf.  From  some  point  o'  in  the 
line  of  action  of  the  left  support  reaction  construct  the  funicular 
of  the  force  diagram  oa'c' .  Draw  oe'  parallel  to  the  closing  line 
e"o'  of  the  funicular.  Then,  e'a'  is  the  magnitude  of  the  left 
support  reaction  Rw'  due  to  the  wind,  and  c'e'  is  the  magnitude 
of  the  right  support  reaction  Rw"  due  to  the  wind.  The  re- 
sultant Ri  of  Rv  and  Rj ',  and  the  resultant  R2  of  Rv  and  Rw" 
are  the  total  support  reactions  in  magnitude  and  direction  due 
to  the  dead  load  and  the  wind  pressure. 

Having  the  magnitudes  and  directions  of  the  resultant  forces 
at  all  the  joints,  the  reciprocal  diagram  can  now  be  drawn. 
Thus: 

Draw  ab,  be,  and  cd  parallel  and  equal  to  the  resultant  forces 
at  joints  2,  3,  and  4.  From  d  drawee  parallel  and  equal  to  R%. 
The  system  being  in  equilibrium  the  force  polygon  must  close; 
therefore  the  space  ea  must  be  exactly  filled  by  a  line  parallel 
and  equal  to  RI',  otherwise  the  construction  would  be  in- 
accurate. 

Commencing  at  the  joint  at  the  left  support  its  reciprocal  eafe 
is  readily  constructed.  Proceeding  then  to  joints  2,  3,  4,  and  5 
the  stresses  in  all  the  members  are  obtained,  the  necessity  of 
the  point  i  falling  on  fe  affording  a  check  as  to  accuracy. 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM 


171 


The  support  reactions  due  to  the  wind  might  have  been  found 
easily  by  moments.  The  frame  diagram  being  a  scale  drawing, 
the  distances  between  the  lines  of  action  of  the  normal  wind 
forces  and  of  the  reactions  due  to  them  are  found  by  measure- 
ment to  be  as  shown  in  Fig.  106.  Then,  taking  moments  about 
the  right  support,  we  have 

26  RJ  =  i  X  1.7  +  4  X  0.86,   whence   Rw'  =  0.82  ton; 
hence,     RJ  =  0.82  X  0.75  =  0.615  inch  =  e'a! . 

Similarly,  by  moments  about  the  left  support,  we  have 

26  RJ'  =  i  X  0.9  +  i  X  1.74,    whence    RJ'  =  0.68  ton; 
hence,     RJ'  =  0.68  X  0.75  =  0.51  inch  =  c'e'. 

The  stresses  in  the  members,  by  scale  measurements  from  the 
reciprocal  diagram,  are  found  to  be  as  here  tabulated: 


Ties 


Struts 


The  solution  of  this  problem  with  the  consideration  of  all  the 
forces,  as  shown  in  Fig.  107,  rejecting  none  at  the  supports,  is 
instructive. 

Lay  off  the  load  line  abcdefghi  to  scale.  Each  of  the  reactions 
Rv  is  equal  to  1.5  tons  =  1X1  =  1  inch  to  scale.  The  effective 
wind  reactions  are  found  by  the  funicular  as  before,  but  the  left 
reaction  thus  found  must  be  augmented  by  the  wind  load  AB 


172          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 


Fig.  107. 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM          173 

in  order  to  obtain  the  full  reaction  Rw'  at  the  left  support  due 
to  the  wind.  Commencing  at  i  the  reactions  RJ' ,  Rv,  RJ ',  Rv 
are  each  set  off  in  magnitude  and  parallel  direction,  closing  the 
force  polygon  at  a. 

The  reciprocal  diagram  can  now  be  drawn,  the  stresses  ob- 
tained being  exactly  the  same  as  by  the  preceding  method.  The 
dotted  lines  aj  and  ji  are  the  resultant  support  reactions  R\  and 
R2  in  magnitude  and  parallel  direction. 

77.  Roof  Truss  Fixed  at  One  End  and  Free  at  the  Other  with 
Wind  Pressure  and  Dead  Vertical  Loads.  —  The  roof  truss  of 
Fig.  1 08  has  a  span  of  50  feet,  and  is  fixed  at  one  end  and  free 
to  move  on  expansion  rollers  at  the  other.  Pitch  of  roof,  28°; 
distance  between  trusses,  16  feet;  dead  load  per  square  foot  of 
horizontal  surface,  17.9  pounds;  wind  pressure  normal  to  the 
truss,  19.77  pounds  per  square  foot.  Linear  scale,  yV  inch  =  i 
foot;  load  scale,  i  inch  =  i  ton.  The  frame  diagram  shows  the 
angular  arrangement  of  the  truss  members,  and  it  will  be  noticed 
that  the  joints  divide  the  rafters  into  three  parts  whose  lengths 
are  to  each  other  as  3  13  :  2. 

The  dead  vertical  load  on  each  truss  =  5°  X  l6  X  I7'9  =  6.4 

2240 

tons,  3.2  tons  on  each  rafter,  and  its  uniform  distribution  places 
0.6  ton  at  joints  i  and  7,  1.2  tons  at  joints  2  and  6,  i  ton  at  joints 
3  and  5,  and  0.8  ton  at  joint  4. 

The  support   reactions   due   to  the  vertical   loads  are  each 

—  =   3.2  tons  =  —  X  -  =  —    inches  when  reduced  to  scale. 

2  IO          2          2O 

These  reactions  are  vertical,  and  each  is  denoted  by  Rv  in  the 
frame  diagram.  The  scale  measurements  of  the  vertical  loads 
at  the  joints  are:  0.6  X  0.5  =  0.3  inch  at  joints  i  and  7; 
1.2  X  0.5  =  0.6  inch  at  joints  2  and  6;  0.5  inch  at  joints  3  and  5; 
and  0.4  inch  at  joint  4.  These  loads  are  laid  off  to  scale  at  the 
joints  in  the  frame  diagram. 


m 

Fig.109 
Scale,  J4  "=1  ton 


d74) 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM          175 
The  total  wind  pressure  to  be  borne  by  one  truss  is 

25  sec  28°  X  16  X  10.77 

2-LI  =  4  tons. 

2240 

This  load  may  act  on  either  side  of  the  truss.  Suppose  first 
that  the  left  end  of  the  truss  be  fixed  and  the  wind  to  act  on  that 
side.  The  distribution  of  the  wind  load  places  0.75  ton  at 
joint  i,  1.5  tons  at  joint  2,  1.25  tons  at  joint  3,  and  0.5  ton  at 
joint  4.  Reduced  to  scale  these  loads  are:  At  joint  i,  0.75 
X  0.5  =  0.375  =  f  inch;  at  joint  2,  1.5  X  0.5  =  J  inch;  at  joint 
3,  1.25  X  0.5  =  0.625  =  f  inch;  and  at  joint  4,  0.5  X  0.5  =  0.25 
=  J  inch.  Set  off  these  loads  to  scale  at  the  joints  in  the  frame 
diagram. 

The  right  end  being  free,  the  reaction  there  due  to  the  wind 
pressure  will  be  vertical. 

Assume  the  truss  to  be  acted  on  only  by  the  wind  and  that  its 
resultant  pressure  of  4  tons  acts  at  the  middle  point  of  the  left 
rafter.  There  are  now  but  three  concurrent  forces,  viz.,  the 
two  support  reactions  and  the  resultant  wind  pressure.  The 
points  of  application  of  these  forces  and  the  directions  of  two 
of  them  being  known,  the  direction  of  the  third  can  be  found. 
Thus: 

Let  fall  a  perpendicular  to  the  left  rafter  from  its  middle 
point,  and  at  the  right  support  let  fall  a  perpendicular  to  the 
bottom  chord  of  the  truss.  These  perpendiculars  are  the  di- 
rections of  the  resultant  normal  wind  pressure  and  of  the 
support  reaction  at  the  free  end  respectively,  and  they  intersect 
at  x.  The  line  joining  x  with  joint  i  will  give  the  direction  of 
the  support  reaction  due  to  the  wind  at  the  left  (fixed)  end.  Lay 
off  to  scale  the  distance  xz  equal  to  the  magnitude,  4  tons,  of 
the  resultant  normal  wind  pressure,  and  resolve  it  into  its  com- 
ponents xy  and  yz  parallel  to  the  support  reactions.  The  reac- 
tion Rwf  at  the  left  support  due  to  the  wind  is  represented  in 


176          THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

magnitude  by  xy.  The  composition  of  RJ  and  Rv  gives  the 
resultant  reaction  RI  at  the  fixed  end  due  to  all  the  external 
forces.  The  addition  of  yz  to  Rv  gives  R2  as  the  support  reac- 
tion at  the  right  (free)  end  due  to  all  the  external  forces.  The 
resultants  of  the  dead  and  wind  loads  at  joints  i,  2,  3,  and  4  are 
found  by  the  parallelogram  of  forces  as  shown. 

The  reciprocal  diagram  for  the  truss  with  its  left  end  fixed 
and  the  wind  blowing  on  the  fixed  side  can  now  be  drawn. 

To  a  scale  of  0.25  inch  to  the  ton  —  one-half  that  used  for 
the  loads  in  the  frame  diagram  —  construct  the  force  polygon 
abcdefghia  of  Fig.  109.  This  polygon  must,  of  course,  close. 
The  diagram  is  readily  completed  by  drawing  in  order  the 
reciprocals  of  the  joints  IABJ,  JBCK,  IJKL,  LKCDM, 
MDEN,  ILMNO,  ONEFP,  and  PFGQ,  the  point  q  falling  at 
the  finish  on  the  line  ij.  The  order  of  procedure  from  joint 
to  joint  determines  itself  from  the  fact  that  the  reciprocal  of 
a  joint  cannot  be  drawn  if  more  than  two  of  the  forces  are 
unknown. 

It  will  be  necessary  to  construct  the  reciprocal  diagram  for 
the  wind  blowing  on  the  free  side  and,  since  the  vertical  loads 
are  the  same  on  each  side,  the  construction  will  be  facilitated  by 
assuming,  in  Fig.  108,  the  right  end  of  the  truss  to  be  fixed  and  the 
left  end  free  to  move  on  rollers. 

The  reaction  at  the  left  end  will  now  be  vertical,  and  its  line 
of  action  intersects  the  line  of  action  of  the  resultant  normal 
wind  pressure  on  the  free  side  at  x',  and  therefore  the  line 
joining  x'  with  joint  7  gives  the  direction  of  the  reaction  at  the 
right  support  due  to  the  wind.  Lay  off  x'z'  equal  to  4  tons  to 
scale  to  represent  the  magnitude  of  the  resultant  wind  pressure. 
Its  component  x'y'  gives  the  magnitude  of  Rj* ',  the  reaction  at 
the  fixed  end  due  to  the  wind.  The  composition  of  Rw"  and  Rv 
gives  RZ  as  the  support  reaction  at  the  right  (fixed)  end  due  to 
all  the  loads.  The  component  y'z'  is  the  amount  to  be  added  to 


FRAMED   STRUCTURES  —  RECIPROCAL  DIAGRAM 


177 


Rv  Sit  the  left  support  to  give  the  total  support  reaction  RI  at 
the  free  end. 

The  force  polygon  and  the  reciprocal  diagram  can  now  be 
drawn  as  shown  in  Fig.  no,  the  point  q  falling  on  the  line  ij. 

The  members  must  be  designed  to  resist  the  maximum  stress 
to  which  they  may  be  subjected,  and  since  the  wind  may  blow 
on  either  side  of  the  truss  they  are  made  the  same  for  each  side. 
The  tabulated  stresses  were  found  by  scale  measurements 
from  Figs.  109  and  no  to  be  as  follows: 

TABLE   OF   STRESSES. 


Members. 

Wind  on  fixed  side. 

Wind  on  free  side. 

Ties. 

Struts. 

Ties. 

Struts. 

BJ 
JI 
CK 
KJ 
KL 
LI 
DM 
ML 
EN 
NM 
NO 
01 
FP 
PO 
GQ 
QP 
QI 

9.20 

9-25 

8.50 
2.6O 

9.60 

7.80 

8.50 
2-55 

2.50 
5-05 

2-55 
6.90 

5.60 
3-40 
5.80 

5.60 
3-40 
5.80 

4-15 
5-9° 

4-05 

i-45 

1-35 

4.00 

7-30 

7-35 

8.00 
i  .10 

I.  00 

1.  10 

7-95 

I.OO 

5-20 

7.00 

78.  A  Framed  Crane.  —  To  draw  the  reciprocal  diagram  of 
the  framed  crane  of  Fig.  in,  having  a  load  of  3  tons  at  its  peak, 
we  proceed  as  follows: 

Letter  the  frame  diagram,  and  then  to  a  scale  of  0.25  inch  = 
i  ton  lay  off  the  load  line  ab  three-fourths  of  an  inch  in  length  to 
represent  the  load  of  3  tons.  From  b  and  a  draw  parallels  to 
BJ  and  JA  respectively.  They  intersect  at  j,  giving  abj  as  the 
force  diagram  of  the  joint  ABJ,  and  bj  and  ja  represent  in 


THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

magnitude  and  direction  the  stresses  in  the  members  BJ  and  JA 
respectively.     Proceed  in  like  manner  to  the  completion  of  the 


reciprocal  diagram  from  which  the  stress  diagram  for  any  joint 
may  at  once  be  read.  Thus,  the  stress  diagram  for  the  joint 
GHBF  is  ghbfg.  The  reaction  R2  is,  of  course,  equal  to  RI  +  W. 


FRAMED   STRUCTURES  —  RECIPROCAL   DIAGRAM          179 

79.  Redundant  and  Deficient  Frames.  —  The  reciprocal  dia- 
grams for  all  the  frames  that  have  been  considered  have  closed, 
and  the  frames,  therefore,  were  complete.  It  will  be  noted  that 
in  the  cantilever  frames  the  number  of  members  equals  twice 
the  number  of  joints  minus  four,  and  that  in  frames  supported 
at  the  ends  the  number  of  members  equals  twice  the  number  of 
joints  minus  three. 

If  a  frame  has  more  than  the  requisite  number  of  members  to 
retain  its  original  shape,  it  is  called  a  redundant  frame  and  is 
statically  indeterminate. 

A  frame  having  an  insufficient  number  of  members  to  pre- 
serve its  original  shape  is  deficient,  but  may  be  used  for  one 
special  distribution  of  the  load.  Any  deviation  from  this  dis- 
tribution will  cause  the  frame  to  change  its  shape.  A  deficient 
frame  may  generally  be  made  complete  by  the  addition  of  another 
member. 


PROBLEMS 

i.  Draw  the  reciprocal  diagram  of  the  Warren  girder  of  Fig.  c,  the 
triangles  being  equilateral.  Span,  42  feet,  and  a  uniformly  distributed 
load  of  1.5  tons  per  foot  is  sustained  by  the  girder.  Tabulate  the  stresses 
and  indicate  their  kinds  in  the  frame  diagram.  Scales:  Load,  o.i  inch  = 
i  ton;  linear,  o.i  inch  =  i  foot. 


2.  Draw  the  reciprocal  diagram  of  the  roof  truss  of  Fig.  d,  which  is 
loaded  as  shown.  Tabulate  the  stresses  and  indicate  their  kinds  in  the 
frame  diagram.  Scales:  Load,  0.25  inch  =  i  ton;  linear,  o.i  inch  =  i 
foot. 


180          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

3 1  tons 

x^s.    /•*  . 

tons 


Fig.  d. 


3.  The  Fink  truss  of  Fig.  e  has  a  span  of  56  feet,  a  depth  of  12  feet, 
and  is  uniformly  loaded  with  1.5  tons  per  foot.  Draw  the  reciprocal  dia- 
gram to  the  scale  of  ^  inch  =  i  ton.  Tabulate  the  stresses  and  indicate 
their  kinds  in  the  frame  diagram.  Scale:  Linear,  ^  inch  =  i  foot. 


Fig.  e. 


4.  The  braced  cantilever  of  Fig.  /  is  20  feet  long,  9  feet  deep,  and  uni- 
formly loaded  on  the  top  with  120  pounds  per  foot.  Draw  the  reciprocal 
diagram  to  the  scale  of  o.i  inch  =  100  pounds.  Tabulate  the  stresses  and 
indicate  their  kinds  in  the  frame  diagram.  Show  that  the  reactions  at  the 
wall  are  equal  but  opposite  in  direction.  Scale:  Linear,  j^  inch  =  i  foot. 


Fig.  /. 


5.  Draw  the  reciprocal  diagram  of  the  braced  cantilever  of  Fig.  g,  and 
tabulate  the  stresses.  Find,  in  magnitude  and  direction,  the  resultant 
stress  on  the  pin  of  the  upper  joint  at  the  wall.  Scales:  Load,  o.i  inch 
=  80  pounds;  linear,  0.25  inch  =  i  foot. 


FRAMED   STRUCTURES  —  RECIPROCAL   DIAGRAM         181 


•Lbs. 


Fig.  g. 

6.  A  roof  truss,  Fig.  h,  of  50  feet  span,  fixed  at  the  ends  and  of  30° 
pitch,  carries  vertical  loads  of  1.5  tons  at  AB,  EC,  CD,  DE  and  EF,  and 
sustains  a  horizontal  wind  pressure  of  33.5  pounds  per  square  foot.  Dis- 
tance between  principals,  12  feet.  Draw  the  reciprocal  diagram  and 
tabulate  the  stresses  in  the  members.  Scales:  Linear,  ^  inch  =  i  foot; 
load,  i  inch  =  i  ton. 


Fig.  h. 

7.  The  roof  truss  of  Fig.  i  has  a  span  of  75  feet  and  a  pitch  of  30°. 
The  rafters  are  divided  into  three  equal  parts  and  the  horizontal  tie  into 
five  equal  parts.  The  distance  between  principals  is  16  feet.  The  dead 
vertical  load  is  6.5  tons,  distributed  as  follows:  At  joints  i  and  7,  0.6  ton 


182 


THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 


each;  at  joint  4,  i  ton;  at  joints  2  and  3,  0.95  ton  each;  at  joints  5  and  6, 
1.2  tons  each.  The  wind  pressure  is  assumed  to  have  a  horizontal  inten- 
sity of  20  pounds  per  square  foot.  Assuming  the  truss  to  be  fixed  at  one 
end  and  free  to  move  on  expansion  rollers  at  the  other,  determine  the 
maximum  stress  in  each  member.  Scales:  Linear,  ^  inch  =i  foot;  load, 
£  inch  =  i  ton. 

8.  The  roof  truss  of  Fig.  j  has  a  span  of  60  feet,  and  supports  dead 
loads  of  2  tons  at  each  of  the  joints  2,  3,  4,  5,  6,  7,  8.  Draw  its  reciprocal 
diagram.  Scales:  Linear,  ^  inch  =  i  foot;  load,  inch  =  i  ton. 


Fig.  j. 


9.  A  bridge  truss,  a  portion  of  which  is  shown  in  Fig.  k,  has  a  span  of 
120  feet  divided  into  six  equal  bays.  It  is  designed  to  carry  a  dead  load 
of  1 6  tons  at  each  of  the  lower  panel  points.  Calculate,  by  the  method  of 
sections,  the  stresses  in  the  members  GM ,  MN,  and  NC,  and  indicate  by 
arrowheads  the  nature  of  the  stresses. 


tons 


Fig.  k. 


FRAMED    STRUCTURES  —  RECIPROCAL   DIAGRAM          183 

10.  Complete  the  frame  diagram  of  problem  9  and  construct  the  re- 
ciprocal diagram,  tabulating  the  magnitudes  and  kinds  of  stresses  in  the 
members.     Scales:  Linear,  ^  inch  =  i  foot;  load,  ^  inch  =  i  ton. 

11.  The  crane  of  Fig.  /  supports  a  load  of  2  tons  at  its  peak.  Deter- 
mine from  the  reciprocal  diagram  the  reactions  and  the  stresses  in  the 
members.     Scales:  Linear,  |  inch  =  i  foot;  load,  i  inch  =  i  ton. 


2  tons 


Fig.  /. 


12.    Calculate  by  the  method  of  sections  the  magnitude  and  kind  of 
stress  in  each  of  the  members  of  the  crane  of  Fig.  m  that  is  cut  by  the  line  xy. 


13.   Draw    the   reciprocal   diagram  of   the  crane  of  Fig.  w.     Scales: 
Linear,  TV  inch  =  i  foot;  load,  ^  inch  =  i  ton. 


CHAPTER  DC 
ENGINEERING   MATERIALS 

80.  Introductory.  —  In  all  engineering  construction  a  knowl- 
edge of  the  properties  of  the  materials  used  is  essential.     It  is 
not    sufficient    to    know  only   the  characteristic  properties  of 
different  materials,  but  their  fitness  for  use  in  any  engineering 
design  must  be  determined  by  actual  test  in  order  to  insure 
safety   to    the    structure,    and    for    this   purpose    the   modern 
mechanical  laboratory  is  equipped  with  special  machines  for 
testing. 

The  materials  used  in  machine  construction  must  be  strong 
and  practically  rigid,  the  element  of  strength  enabling  the  size 
and  weight  of  machines  to  be  reduced  to  a  minimum,  and  the 
rigidity  enabling  the  material  to  resist  the  tendency  to  change 
its  size  or  shape  while  under  the  influence  of  straining  actions. 
As  no  materials  are  absolutely  rigid,  it  follows  that  the  materials 
of  machines  undergo  some  deformation  under  the  action  of  the 
forces  to  which  they  are  subjected,  but  by  judicious  selection  of 
materials  and  a  proper  proportioning  of  parts,  such  deforma- 
tions may  be  rendered  so  small  as  to  be  negligible.  Generally 
speaking,  the  deformations  in  machines  are  temporary  in  char- 
acter, the  elasticity  of  the  chosen  materials  causing  them  to 
vanish  upon  the  removal  of  the  straining  force. 

The  principal  engineering  materials  are  cast  iron,  wrought  iron, 
steel,  copper,  lead,  tin,  zinc,  alloys,  timber,  concrete,  and  reinforced 
concrete. 

81.  CAST  IRON.  —  Cast  iron  is  obtained  in  the  form  of  pig 

iron  by  direct  treatment  of  the  ore  in  a  blast  furnace,  its  quality 

184 


ENGINEERING  MATERIALS  185 

depending  on  the  relative  amounts  of  other  substances  in  its 
combination. 

Influence  of  Carbon.  —  In  the  molten  condition  the  carbon  is 
dissolved  by  the  iron  and  held  in  solution  just  as  ordinary  salt 
is  dissolved  by  water.  The  mixture  or  combination  of  the  two 
elements  is  thus  entirely  uniform.  The  proportion  of  carbon 
which  pure  melted  iron  can  thus  dissolve  and  hold  in  solution 
is  about  3.5  per  cent.  If  chromium  or  manganese  is  present  also, 
the  capacity  for  carbon  is  much  increased,  while  with  silicon,  on 
the  other  hand,  the  capacity  for  carbon  is  decreased.  In  the 
various  grades  of  cast  iron  the  proportion  of  carbon  is  usually 
found  to  be  between  2  per  cent  and  4.5  per  cent. 

When  such  a  molten  mixture  cools  and  becomes  solid  there  is 
a  tendency  for  a  part  of  the  carbon  to  be  separated  and  no  longer 
remain  in  intimate  combination  with  the  iron.  The  carbon  thus 
separated,  or  precipitated,  from  the  iron  takes  that  form  known 
as  graphite  and  collects  in  very  small  flakes  or  scales.  The 
carbon  which  remains  in  intimate  combination  with  the  iron  is 
said  to  be  combined,  while  that  which  is  separated  is  usually 
called  graphitic. 

The  qualities  of  cast  iron  depend  chiefly  on  the  proportion  of 
total  carbon  and  on  the  relative  proportion  of  combined  and 
graphitic  carbon. 

With  a  high  proportion  of  graphitic  carbon  the  iron  is  soft  and 
tough,  with  a  low  tensile  strength,  and  breaks  with  a  dark  and 
coarse-grained  fracture.  In  fact,  the  substance  in  this  condition 
may  be  considered  as  nearly  pure  iron  with  fine  flakes  of  graphite 
entangled  and  distributed  through  it,  thus  giving  to  the  iron  a 
spongy  structure.  The  iron  thus  forms  a  kind  of  continuous 
mesh  about  the  graphite,  which  decreases  the  strength  by  reason 
of  the  decrease  of  cross-sectional  area  actually  occupied  by  the 
iron  itself.  Such  irons  are  termed  gray. 

As  the  relative  proportion  of  graphitic  carbon  decreases  and 


1 86          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

that  of  combined  carbon  increases,  the  iron  takes  on  new  prop- 
erties, becoming  harder  and  more  brittle.  Its  tensile  strength 
also  increases  to  a  certain  extent,  and  the  fracture  becomes  fine- 
grained, or  smooth,  and  whiter  in  color.  When  these  charac- 
teristics are  pronounced,  the  iron  is  said  to  be  white.  When 
about  half  the  carbon  is  combined  and  half  separates  as  graphite 
the  effect  is  to  produce  a  distribution  of  dark  spots  scattered 
over  a  whitish  field.  Such  iron  is  said  to  be  mottled. 

In  a  general  way,  with  a  large  proportion  of  total  carbon 
there  is  likely  to  be  formed  a  considerable  amount  of  graphitic 
carbon,  and  hence  such  iron  is  usually  gray  and  soft.  Also, 
with  a  large  proportion  of  carbon  the  iron  melts  more  readily 
and  its  fluidity  is  more  pronounced.  As  the  proportion  of  total 
carbon  decreases  the  cast  iron  gradually  approaches  the  con- 
dition of  steel,  whose  properties  will  be  discussed  in  later 
paragraphs. 

Of  the  special  ingredients  in  cast  iron  the  combined  carbon  is 
the  one  of  greatest  importance.  It  is  that  chiefly  which,  by  unit- 
ing with  the  iron,  gives  it  new  qualities,  and  the  principal  influ- 
ence of  other  substances  lies  in  the  effect  which  they  may  have 
on  the  proportion  of  this  ingredient.  As  between  graphitic 
and  combined  carbon  the  former  does  not  affect  the  quality 
of  the  iron  itself,  but  acts  physically  by  affecting  the  structure 
of  the  casting,  while  the  latter  by  entering  into  combination  with 
the  iron  acts  chemically,  and  produces  a  new  substance  with 
different  qualities. 

The  proportions  of  combined  and  graphitic  carbon  are  influ- 
enced by  the  rate  of  cooling  and  by  the  presence  or  absence  of 
various  other  ingredients.  Slow  cooling  allows  time  for  the 
separation  of  the  carbon  and  thus  tends  to  form  graphitic  carbon 
and  soft  gray  irons.  Quick  cooling,  or  chilling  in  the  extreme 
case,  prevents  the  formation  of  graphitic  carbon  and  thus  tends 
to  form  hard,  white  iron. 


ENGINEERING   MATERIALS  187 

In  addition  to  carbon,  small  particles  of  silicon,  sulphur,  phos- 
phorus, manganese,  and  chromium  may  be  found  in  cast  iron. 

Influence  of  Silicon.  —  The  fundamental  influences  of  silicon 
are  two.  (a)  It  tends  to  expel  the  carbon  from  the  combined 
state  and  thus  to  decrease  the  relative  proportion  of  combined 
carbon  and  increase  that  of  graphitic  carbon,  (b)  Of  itself, 
silicon  tends  to  harden  cast  iron  and  to  make  it  brittle.  These 
two  influences  are  opposite  in  character,  since  an  increase  in 
graphitic  carbon  softens  the  iron.  In  usual  cases  the  net  result 
is  a  softening  of  the  iron,  an  increase  in  fluidity,  and  a  general 
change  toward  those  qualities  possessed  by  iron  with  a  high 
proportion  of  graphitic  carbon.  This  applies  with  a  proportion 
of  silicon  from  2  per  cent  to  4  per  cent.  With  more  than  this 
the  influence  on  the  carbon  is  but  slight  and  the  result  on  the 
iron  is  to  decrease  the  strength  and  toughness,  giving  a  hard  but 
brittle  and  weak  grade  of  iron. 

A  chilled  cast  iron  is  an  iron  which,  if  cooled  slowly,  would  be 
gray  and  soft,  but  if  cooled  suddenly  by  contact  with  a  metal 
mold,  or  by  other  means,  becomes  white  or  hard,  especially  at 
and  near  the  surface.  Certain  grades  of  cast  iron  tend  to  chill 
when  cast  in  sand  molds.  This  property  is  usually  undesirable. 
In  such  cases  the  tendency  may  be  prevented  by  the  addition  of 
silicon,  which,  by  forcing  the  carbon  into  the  graphitic  state  on 
cooling,  prevents  the  formation  of  hard,  chilled  surfaces.  In  all 
cases  the  actual  effect  of  adding  silicon  will  depend  much  on  the 
character  of  the  iron  used  as  a  base,  and  only  a  statement  of  the 
general  tendencies  can  here  be  given. 

To  sum  up,  a  white  iron  which  would  give  hard,  brittle,  and 
porous  castings  can  be  made  softer,  tougher,  and  more  solid  by 
the  addition  of  silicon  to  the  extent  of  perhaps  2  per  cent  or  3 
per  cent.  As  the  silicon  is  increased,  the  iron  will  become  softer 
and  grayer  and  the  tensile  strength  will  decrease.  At  the  same 
time  the  shrinkage  will  decrease,  at  least  for  a  time,  though  it 


1 88          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

may  increase  again  with  large  excess  of  silicon.  The  softening 
and  toughening  influence,  however,  will  only  continue  so  long  as 
additional  graphite  is  formed,  and  when  most  of  the  carbon  is 
brought  into  this  state  the  maximum  effect  has  been  produced 
and  any  further  addition  of  silicon  will  decrease  both  strength 
and  toughness. 

Influence  of  Sulphur.  —  Authorities  are  not  in  entire  agree- 
ment as  to  the  influence  of  sulphur  on  cast  iron,  some  believing 
that  it  tends  to  increase  the  proportion  of  combined  carbon, 
while  others  maintain  that  it  tends  to  decrease  both  the  com- 
bined carbon  and  the  silicon.  It  is  generally  agreed,  however, 
that  in  proportions  greater  than  about  0.15  to  0.20  of  i  per 
cent  it  increases  the  shrinkage  and  the  tendency  to  chill  and 
decreases  the  strength.  Sulphur  does  not  readily  enter  cast  iron 
under  ordinary  conditions  and  its  influence  is  not  especially 
feared.  An  increase  in  the  proportion  of  sulphur  in  cast  iron  is 
most  likely  to  result  from  an  absorption  of  sulphur  in  the  coke 
during  the  operation  of  melting  in  the  cupola. 

Influence  of  Phosphorus.  —  The  presence  of  phosphorus  in- 
creases the  fusibility,  fluidity,  and  brittleness  of  cast  iron  and  is 
desirable  in  light  and  ornamental  castings.  The  maximum 
amount  of  phosphorus  should  not  exceed  1.5  per  cent. 

Influence  of  Manganese.  —  This  element  by  itself  decreases 
fluidity,  increases  shrinkage,  and  makes  the  iron  harder  and  more 
brittle.  It  combines  with  iron  in  all  proportions.  The  com- 
bination containing  less  than  50  per  cent  of  manganese  is  called 
spiegeleisen.  With  more  than  50  per  cent  of  manganese  it  is 
called  ferromanganese.  One  of  the  most  important  properties  of 
manganese  in  combination  with  iron  is  that  it  increases  the 
capacity  of  the  iron  for  carbon.  Pure  iron  will  take  only  about 
3.5  per  cent  of  carbon,  while  with  the  addition  of  manganese  the 
proportion  may  rise  to  6  or  7  per  cent.  Manganese  is  also  be- 
lieved to  decrease  the  capacity  of  iron  for  sulphur  and  to  this 


ENGINEERING  MATERIALS  189 

extent  may  be  a  desirable  ingredient  in  proportions  not  exceed- 
ing i  to  1.5  per  cent. 

Shrinkage.  —  At  the  moment  of  hardening,  cast  iron  expands 
and  takes  a  good  impression  of  the  mold.  In  the  gradual  cooling 
after  setting,  however,  the  metal  contracts,  so  that  on  the  whole 
there  is  a  shrinkage  of  about  0.125  inch  per  foot  in  all  directions, 
though  this  amount  varies  somewhat  with  the  quality  of  the 
iron  and  with  the  form  and  dimensions  of  the  pattern.  In  a 
general  way,  hardness  and  shrinkage  increase  and  decrease 
together. 

Strength  and  Hardness.  —  The  strength  of  cast  iron  is  chiefly 
dependent  upon  its  amount  of  combined  carbon.  The  greatest 
crushing  strength  is  obtained  with  sufficient  combined  carbon 
to  make  a  rather  hard,  white  iron,  while  for  the  maximum 
transverse  or  bending  strength  the  combined  carbon  is  some- 
what less  and  the  iron  only  moderately  hard.  For  the  greatest 
tensile  strength,  the  combined  carbon  is  still  less  and  the  iron 
rather  soft.  Metal  still  softer  than  this  works  with  the  greatest 
facility  but  is  deficient  in  strength. 

Uses  in  Engineering.  —  Cast  iron  is  used  for  cylinders,  cylin- 
der heads,  liners,  slide  valves,  valve  chests  and  connections,  and 
generally  for  all  parts  having  considerable  complexity  of  form. 
It  is  also  used  for  columns,  bed  plates,  bearing  pedestals,  caps, 
etc.,  though  cast  and  forged  steel  are  to  some  extent  displacing 
cast  iron  for  some  of  these  uses.  It  is  also  used  for  grate  bars, 
furnace  door  frames,  and  for  minor  boiler  fittings. 

Inspection  of  Castings.  —  In  the  inspection  of  castings  care 
must  be  had  to  note  the  texture  of  the  surface  and  to  this  end 
the  outer  scale  and  burnt  sand  should  be  carefully  removed  by 
the  use  of  brushes  or  chipping  hammer,  or,  if  necessary,  by 
pickling  in  dilute  muriatic  acid.  The  flaws  most  liable  to  occur 
are  blow  holes  and  shrinkage  cracks.  The  parts  of  the  casting 
most  liable  to  be  affected  by  blow  holes  are  those  on  the  upper 


I  go          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

side  or  near  the  top,  and  on  this  account  a  sinking  head  or  extra 
piece  is  often  cast  on  top,  into  which  the  gases  and  impurities 
may  collect.  This  is  afterwards  cut  off,  leaving  the  sounder 
metal  below.  Shrinkage  cracks  are  of  unusual  occurrence.  The 
presence  of  blow  holes,  if  large  in  size  or  in  great  number  and 
near  the  surface,  may  often  be  determined  by  tapping  with  a 
hammer.  The  sound  given  out  will  serve  to  indicate  to  an  ex- 
perienced ear  the  probable  character  of  the  metal  underneath. 

Brazing  of  Cast  Iron.  —  Cast  iron  may  be  brazed  to  itself,  or 
to  most  of  the  structural  metals,  by  the  use  of  a  brazing  solder 
of  suitable  melting  point  and  with  proper  care  in  the  operation. 
Cast  iron  may  also  be  united  to  itself,  or  to  wrought  iron  or 
steel,  by  the  operation  of  burning.  This  consists  in  placing  in 
position  the  two  pieces  to  be  united,  and  then  allowing  a  stream 
of  melted  cast  iron  to  flow  over  the  surfaces  to  be  joined,  the 
adjacent  parts  being  protected  by  fire  clay  or  other  suitable 
material.  The  result  is  to  soften  or  partially  melt  the  surfaces 
of  the  pieces,  and  by  arresting  the  operation  at  the  right  moment 
they  may  be  securely  joined  together. 

Malleable  Cast  Iron.  —  If  iron  castings  of  not  too  great  thick- 
ness, and  of  such  purity  as  to  be  low  in  sulphur,  be  embedded  in 
powdered  red  oxide  of  iron  (red  hematite)  and  maintained  at  a 
red  heat  for  two  or  three  days,  they  become,  in  a  measure,  de- 
carbonized as  a  result  of  the  chemical  action  which  ensues. 
The  carbon  first  disappears  from  the  outer  layers,  and  as  the 
process  continues  decarbonization  toward  the  innermost  layers 
takes  place.  If  the  process  be  carried  to  the  extreme  in  the 
effort  to  withdraw  all  the  carbon  from  the  interior  the  outer 
layer  is  very  liable  to  become  brittle,  thus  defeating  the  object 
to  be  attained.  For  this  reason  there  always  remains  a  core  of 
cast  iron  only  partially  decarbonized,  while  the  outermost  layers 
are  left  in  the  condition  of  soft  or  malleable  iron.  The  process 
has  little  effect  upon  the  sulphur,  manganese,  phosphorus,  and 


ENGINEERING  MATERIALS  IQI 

other  impurities  of  the  castings,  but  the  resulting  product  is  a 
malleable  casting  much  less  fusible  than  cast  iron  and  possessing 
six  times  its  ductility.  The  best  product  may  be  twisted  and 
bent  to  a  considerable  extent  before  breaking,  and  its  ability  to 
withstand  shocks  is  much  greater  than  that  of  cast  iron.  Pipe 
fittings,  to  some  extent,  are  malleable  castings,  and  this  mate- 
rial is  largely  used  in  appliances  of  car  construction  where  more 
strength  and  toughness  are  required  than  cast  iron  affords. 

82.  WROUGHT  IRON.  —  Wrought  iron  is  nearly  pure  iron 
mixed  with  more  or  less  slag.  Nearly  all  the  wrought  iron  used 
in  modern  times  is  made  from  cast  iron  by  the  puddling  process 
in  a  reverberatory  furnace.  For  the  details  of  this  process 
reference  may  be  had  to  textbooks  on  metallurgy.  We  can  only 
note  here  that,  in  a  furnace  somewhat  similar  to  the  open  hearth, 
most  of  the  carbon,  silicon,  and  other  special  ingredients  of 
cast  iron  are  removed  by  the  combined  action  of  the  flame  and 
of  a  molten  bath  of  slag  or  fluxing  material,  consisting  chiefly  of 
black  oxide  of  iron.  As  this  process  approaches  completion 
small  bits  of  nearly  pure  iron  separate  from  the  bath  of  melted 
slag  and  unite.  This  operation  is  assisted  by  the  puddling  bar, 
and  after  the  iron  has  thus  become  separated  from  the  liquid  slag 
it  is  taken  out,  hammered  or  squeezed,  and  rolled  into  bars  or 
plates.  Some  of  the  slag  is  necessarily  retained  in  the  iron  and, 
by  the  process  of  manufacture,  is  drawn  out  into  fine  threads, 
giving  to  the  iron  a  stringy  or  fibrous  appearance  when  nicked 
and  bent  over  or  pulled  apart. 

The  proportion  of  carbon  in  wrought  iron  is  very  small,  rang- 
ing from  0.02  to  0.20  of  i  per  cent.  In  addition,  small  amounts 
of  sulphur,  phosphorus,  silicon,  and  manganese  are  usually 
present. 

The  proportion  of  sulphur  should  not  exceed  o.oi  of  i  per  cent. 
Excess  of  sulphur  makes  the  iron  red-short,  that  is,  brittle  when 
red  hot. 


IQ2          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

The  proportion  of  phosphorus  may  vary  from  0.05  to  0.25  of 
i  per  cent.  Excess  of  phosphorus  makes  the  metal  cold-short, 
that  is,  brittle  when  cold. 

The  proportion  of  silicon  may  vary  from  0.05  to  0.30  of  i  per 
cent. 

The  proportion  of  manganese  may  vary  from  0.005  to  0.05  of 
i  per  cent.  The  influence  of  the  silicon  and  manganese  is  usu- 
ally slight  and  unimportant. 

^  Special  Properties.  —  Wrought  iron  is  malleable  and  ductile, 
and  may  be  rolled,  flanged,  forged,  and  welded.  It  cannot  be 
hardened,  though  by  the  process  of  case-hardening  a  surface 
layer  of  steel  is  formed  which  may  be  hardened.  Wrought  iron 
may  be  welded,  because  for  a  considerable  range  of  temperature 
below  melting  (which  takes  place  only  at  a  very  high  tempera- 
ture) the  iron  becomes  soft  and  plastic,  and  two  pieces  pressed 
together  in  this  condition  unite  and  form,  on  cooling,  a  junction 
nearly  as  strong  as  the  solid  metal.  In  order  that  this  welding 
operation  may  be  successful,  the  iron  must  be  heated  sufficiently 
to  bring  it  to  the  plastic  condition,  yet  not  overheated,  and 
there  must  be  employed  a  flux  (usually  borax)  which  will  unite 
with  the  iron  oxide  and  other  impurities  at  the  joint  and  form  a 
thin  liquid  slag  which  may  readily  be  pressed  out  in  the  oper- 
ation, thus  allowing  the  clean  metal  surfaces  of  the  iron  to  effect 
a  union  as  desired. 

83.  STEEL.  — -  Steel  may  be  made  from  wrought  iron  by  in- 
creasing its  proportion  of  carbon,  or  from  cast  iron  by  decreas- 
ing its  proportion  of  carbon.  The  earlier  processes  followed  the 
first  method,  and  high-grade  steels  are  still  made  in  this  way  by 
the  crucible  process. 

The  properties  of  steel  depend  partly  on  the  proportions  of 
carbon  and  other  ingredients  which  it  may  contain,  and  partly 
on  the  process  of  manufacture.  The  proportion  of  carbon  is 
intermediate  between  that  for  wrought  iron  and  for  cast  iron. 


ENGINEERING   MATERIALS  193 

In  the  so-called  mild  or  structural  steel  the  carbon  is  usually 
from  o.i  to  0.3  of  i  per  cent.  In  spring  steel  the  carbon  propor- 
tion is  somewhat  greater,  and  in  high  carbon  grades,  such  as  are 
used  for  tools,  etc.,  the  carbon  is  from  0.6  to  1.2  per  cent.  In 
addition  to  the  carbon  there  may  be  sulphur,  phosphorus,  silicon, 
and  manganese  in  varying  but  very  small  amounts. 

Crucible  Steel.  —  In  the  crucible  process  of  making  steel  a 
pure  grade  of  wrought  iron  is  rolled  into  flat  bars.  These  are 
then  cut,  piled,  and  packed  with  intermediate  layers  of  charcoal 
and  subjected  to  a  high  temperature  for  several  days.  This 
recarbonizes  or  adds  carbon  to  the  wrought  iron  and  thus  makes 
what  is  called  cement  or  blister  steel.  These  bars  are  then  broken 
into  pieces  of  convenient  size,  placed  in  small  crucibles,  melted, 
and  cast  into  bars,  or  into  such  shapes  as  are  desired. 

Structural  Steel.  —  Structural  or  mild  steel  is  made  by  the 
second  general  process,  that  of  reducing  the  proportion  of  carbon 
in  cast  iron.  In  this  operation  there  are  two  processes,  known 
as  the  Bessemer  and  the  Siemens-Martin  or  open  hearth. 

Bessemer  Process.  —  In  this  process  the  carbon  and  silicon 
are  burned  almost  entirely  out  of  the  cast  iron  by  forcing  an 
air  blast  through  the  molten  iron  in  a  vessel  known  as  a  con- 
verter. A  small  amount  of  spiegeleisen,  or  iron  rich  in  carbon 
and  manganese,  is  then  added  in  such  quantity  as  to  make  the 
proportion  of  carbon  and  manganese  suitable  for  the  charge  as 
a  whole.  The  steel  thus  formed  is  then  cast  into  ingots,  or  into 
such  forms  as  may  be  desired.  In  this  process  no  sulphur  or 
phosphorus  is  removed,  so  that  it  is  necessary  to  use  a  cast  iron 
nearly  free  from  these  ingredients  in  order  that  the  steel  may 
have  the  properties  desired. 

A  modification,  by  means  of  which  the  phosphorus  is  removed, 
and  known  as  the  basic  Bessemer  process,  is  used  to  some  ex- 
tent. In  this  process,  calcined  or  burnt  lime  is  added  to  the 
charge  just  before  pouring.  This  unites  with  the  phosphorus, 


194          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

removes  it  from  the  steel,  and  brings  it  into  the  slag.  In  the 
basic  process  the  converter  is  lined  with  ganister  —  a  mixture 
of  ground  quartz  and  fire  clay  —  to  protect  it  from  attack  by 
the  limestone  added  to  the  charge  and  from  the  resulting  slag. 

In  the  Bessemer  process  first  noted,  often  known  as  the  acid 
process,  in  distinction  from  the  basic  process,  the  lining  of  the 
converter  is  of  ordinary  fire  clay. 

The  removal  of  the  phosphorus  by  the  basic  process  makes 
possible  the  use  of  an  inferior  grade  of  cast  iron.  At  the  same 
time,  engineers  are  not  altogether  agreed  as  to  the  relative 
values  of  the  two  products,  and  many  prefer  steel  made  by  the 
acid  process  from  an  iron  nearly  free  from  phosphorus  at  the 
start. 

The  Open-hearth  Process.  —  In  this  process  a  charge  of 
material  consisting  of  wrought  iron,  cast  iron,  steel  scrap,  and 
sometimes  certain  ores,  is  melted  on  the  hearth  of  a  reverbera- 
tory  furnace  heated  by  gas  fuel  on  the  Siemens-Martin  or  regen- 
erative system.  The  carbon  is  thus  partially  burnt  out  in  much 
the  same  manner  as  for  wrought  iron,  and  the  proportion  of 
carbon  is  brought  down  to  the  desired  point,  or  slightly  below 
that  point.  A  charge  of  spiegeleisen  is  then  added  in  order  that 
the  manganese  may  act  on  any  oxide  of  iron  slag  which  remains 
in  the  bath  which,  if  allowed  to  form  a  part  of  the  charge,  would 
make  the  steel  red-short.  The  manganese  separates  the  iron 
from  the  oxide  and  returns  it  to  the  bath,  while  the  carbon  joins 
with  that  already  present  and  thus  produces  the  desired  propor- 
tions. 

Here,  as  with  the  similar  operations  with  the  Bessemer  con- 
verter, there  is  no  removal  of  sulphur  or  of  phosphorus,  and  only 
materials  nearly  free  from  these  ingredients  can  be  used  for 
steel  of  satisfactory  quality.  With  very  low  carbon,  however,  a 
little  phosphorus  seems  to  be  desirable  to  add  strength  to  the 
metal.  This  limitation  of  the  available  materials  has  led,  as  with 


ENGINEERING   MATERIALS  195 

the  Bessemer  process,  to  the  use  of  calcined  limestone  in  the 
charge,  its  purpose  being  to  unite  with  most  of  the  phosphorus 
and  hold  it  in  the  slag.  As  with  the  Bessemer  process,  it  is 
necessary  in  this  case  to  use  a  basic  lining  for  the  furnace,  and 
it  is  known  as  the  basic  open-hearth  process.  The  process 
which  does  not  use  limestone  in  the  charge  has  come  to  be 
known  as  the  acid  open-hearth  process.  There  is  much  differ- 
ence of  opinion  as  to  the  relative  merits  of  the  two  open-hearth 
processes.  Either  will  produce  good  steel  with  proper  care  and 
neither  will  without  it.  It  is  usually  considered  sufficient  to 
specify  the  allowable  limits  of  phosphorus  and  sulphur  and 
then  leave  the  choice  of  the  acid  or  basic  process  to  the 
maker. 

Open-hearth  and  Bessemer  Steels  Compared.  —  Open-hearth 
steel  is  usually  preferred  for  structural  material,  for  these  reasons: 

(a)  It  seems  to  be  more  reliable  and  less  subject  to  unexpected 
or  unexplained  failure  than  the  Bessemer  product. 

(b)  Analysis  shows  that  it  is  much  more   homogeneous  in 
compositign  than  Bessemer  steel,  and  experience  shows  that  it 
is  much  more  uniform  in  physical  quality.     This  is  due  to  the 
process  of  manufacture,  which  is  much  more  favorable  to  a 
thorough  mixing  of  the  charge  than  in  the  Bessemer  process. 

(c)  The  open-hearth  steel  may  be  tested  from  time  to  time 
during  the  operation,  so  that  its  composition  may  be  determined 
and  adjusted  to  fulfill  special  conditions.     This  is  not  possible 
with  the  Bessemer  process. 

Influence  of  Sulphur.  —  Sulphur  makes  steel  red-short  and 
interferes  with  its  forging  and  welding  properties.  Manganese 
tends  to  counteract  the  bad  effects  of  sulphur.  Good  crucible 
steel  has  rarely  more  than  o.oi  of  i  per  cent  of  sulphur.  In 
structural  steel  the  proportion  may  vary  from  0.02  to  o.-io  of  i 
per  cent.  When  possible  it  should  be  reduced  to  not  more  than 
0.03  or  0.04  of  i  per  cent. 


196          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

Influence  of  Phosphorus.  —  Phosphorus  increases  the  tensile 
strength  and  raises  the  elastic  limit  of  low  carbon  or  structural 
steel,  but  at  the  expense  of  its  ductility  and  toughness,  or  ability 
to  withstand  shocks  and  irregularly  applied  loads.  It  is  thus 
considered  as  a  dangerous  ingredient  and  the  amount  allowable 
should  be  carefully  specified.  This  is  usually  placed  from  0.02 
to  o.io  of  i  per  cent. 

Influence  of  Silicon.  —  Silicon  tends  to  increase  the  tensile 
strength  and  to  reduce  the  ductility  of  steel.  It  also  increases  the 
soundness  of  castings  and  ingots  and,  by  reducing  the  iron  oxide, 
tends  to  prevent  red-shortness.  The  process  of  manufacture 
usually  removes  nearly  all  of  the  silicon,  so  that  it  is  not  an  ele- 
ment likely  to  give  trouble  to  the  steel  makers.  The  proportion 
allowable  should  not  be  more  than  from  o.i  to  0.2  of  i  per 
cent. 

Influence  of  Manganese.  —  This  element  is  believed  to  in- 
crease hardness  and  fluidity,  and  to  raise  the  elastic  limit  and 
increase  the  tensile  strength.  It  also  removes  iron  oxide  and 
sulphur  and  tends  to  counteract  the  influence  of  such  amounts 
of  sulphur  and  phosphorus  as  may  remain.  It  is  thus  an  im- 
portant factor  in  preventing  red-shortness.  The  proportion 
needed  to  obtain  these  valuable  effects  is  usually  found  between 
0.2  and  0.5  of  i  per  cent. 

Semisteel.  —  A  metal  bearing  this  trade  name  has  in  recent 
years  attracted  favorable  attention  and  has  come  into  consid- 
erable use  where  somewhat  greater  strength  and  toughness  are 
required  than  can  be  provided  by  cast  iron.  It  is  made  by 
melting  mild  steel  scrap,  such  as  punchings  and  clippings  of 
boiler  plate,  with  cast-iron  pig  in  the  proportions  of  25  or  30 
per  cent  of  the  former  to  75  or  70  per  cent  of  the  latter.  The 
preserice  of  manganese  and  other  special  fluxes  in  small  pro- 
portions is  found  to  add  essentially  to  the  strength,  toughness, 
and  ability  to  withstand  shocks  decidedly  greater  than  for  cast 


ENGINEERING   MATERIALS  1 97 

iron,  and  with  fairly  good  machine  qualities.  Semisteel  casts 
as  readily  as  most  grades  of  cast  iron,  and  its  shrinkage  and 
general  manipulation  are  about  the  same. 

Mechanical  Properties.  —  The  tensile  strength  of  the  lowest 
carbon  steel,  say  about  o.io  of  i  per  cent  carbon,  is  usually 
from  50,000  to  55,000  pounds  per  square  inch  of  section.  The 
strength  increases  quite  uniformly  with  the  increase  of  carbon, 
provided  there  are  no  unusual  proportions  of  sulphur  and  phos- 
phorus. Experiment  shows  that  under  these  circumstances  the 
tensile  strength  will  increase  up  to  75,000  pounds  per  square 
inch,  or  higher,  at  the  rate  of  from  1200  to  1500  pounds  per  o.oi 
of  i  per  cent  of  carbon  added.  At  the  same  time,  with  the  in- 
crease in  strength  the  ductility  decreases,  so  that  a  proper  choice 
must  be  made  according  to  the  particular  uses  for  which  the 
steel  is  intended.  In  the  best  grades  of  tool  steel,  with  carbon 
ranging  from  0.5  to  i.o  per  cent  or  over,  the  strength  ranges 
from  80,000  to  120,000  pounds,  and  even  higher  in  exceptional 
cases. 

Special  Properties.  —  Mild  or  low  carbon  steel  may  be  welded, 
forged,  flanged,  rolled,  and  cast.  It  cannot  be  tempered  or 
hardened  with  a  proportion  of  carbon  lower  than  about  0.75  of 
i  per  cent.  High  carbon  steel  can  be  welded  only  imperfectly 
and,  if  very  high  in  carbon,  not  at  all.  It  can  be  forged  with 
care  and  cast  in  forms  as  desired.  It  can  be  tempered  or  hard- 
ened by  heating  to  a  full  yellow  and  quenching  in  cold  water  or  by 
other  means,  and  then  drawing  the  temper  to  the  point  desired. 

Mild  steel  should  not  be  worked  under  the  hammer  or  flanging 
press  at  a  low,  or  blue,  heat,  as  such  working  is  found  in  many 
cases  to  leave  the  metal  brittle  and  unreliable.  Steel,  in  order 
to  weld  satisfactorily,  should  have  a  low  proportion  of  sulphur, 
and  special  care  is  required  in  the  operation,  because  the  range 
of  temperature  through  which  the  metal  is  plastic  and  fit  for 
welding  is  less  than  with  wrought  iron. 


198          THE  ELEMENTS  OF  MECHANICS  OF    MATERIALS 

Tempering.  —  In  the  operation  of  tempering,  the  steel  after 
quenching  is  very  hard  and  brittle.  In  order  to  give  the  metal 
the  properties  desired,  the  temper  is  drawn  down  by  reheating 
it  to  a  certain  temperature  and  then  quenching  again;  or  better 
still,  by  allowing  it  to  cool  gradually,  provided  the  temper  does 
not  rise  above  the  limiting  value  suitable  for  the  purpose  de- 
sired. If  the  reheating  is  done  in  a  bath  of  oil,  the  conditions 
may  be  kept  under  good  control  and  the  final  cooling  may  be 
slow.  If  the  reheating  is  in  or  over  a  fire,  the  control  is  lacking 
and  the  piece  must  be  quenched  as  soon  as  the  proper  tempera- 
ture is  reached.  This  is  usually  determined  by  the  color  of  the 
oxide  or  scale  that  forms  on  the  brightened  surface  of  the  metal. 
The  following  table  shows  the  temperatures,  the  corresponding 
colors,  and  the  uses  for  which  the  different  tempers  are  suited: 

430°  Faint  yellow 


450°  Straw  yellow 


Hardest  and  keenest  cutting  tools. 


470°  Full  yellow 

490°  Brown  yellow  1  Cutting  tools  requiring  less  hardness 

or  orange    J     and  more  toughness. 
510°  Purplish  )  Tools  for  softer  materials,  or  those 

530°  Purple  J     required  to  stand  rough  usage. 


550°  Light  blue 
560°  Full  blue 
600°  Dark  blue 


Spring  temper.  Used  for  tools  re- 
quiring great  elasticity,  and  those 
for  working  very  soft  materials. 


Special  Steels.  —  In  the  common  grades  of  steel  the  valuable 
properties  are  due  to  the  presence  of  carbon,  modified  in  some 
degree  by  other  ingredients.  There  are  other  substances  which, 
when  united  with  iron  in  small  proportions,  give  to  the  combina- 
tion increased  strength,  hardness,  or  other  valuable  properties. 
We  have  thus  various  special  steels  in  which  their  properties 
may  be  due  to  the  presence  of  carbon  and  other  ingredients,  or 
due  chiefly  to  special  ingredients  other  than  carbon.  The  most 


ENGINEERING  MATERIALS  1 99 

important  of  the  special  steels  are  known  as  nickel  steel  and 
tungsten  steel. 

Nickel  Steel.  —  An  alloy  known  as  nickel  steel,  containing 
about  3  per  cent  of  nickel  and  varying  amounts  of  carbon,  is 
found  to  have  increased  strength  and  toughness  as  compared 
with  ordinary  steel.  It  is  extensively  used  in  the  manufacture 
of  guns  and  armor  plate,  and  to  some  extent  it  has  been  em- 
ployed in  government  work  for  propeller  shafts  and  for  boiler 
plates. 

Tungsten  Steel.  —  This  steel,  known  also  as  Musket  steel, 
containing  tungsten  in  proportions  varying  from  8  to  15  per 
cent,  is  very  hard  and  can  be  forged  only  by  the  exercise  of 
great  care.  Its  hardness  is  not  increased  by  tempering  but  is 
naturally  acquired  as  the  metal  cools;  hence  it  is  said  to  be 
self-hardening.  Some  specimens  contain  also  small  amounts  of 
manganese  and  silver.  Its  chief  use  is  for  lathe,  planer,  and 
other  cutting  and  shearing  tools  where  excessive  hardness  is 
required. 

Uses  in  Engineering.  —  Structural  steel  is  used  extensively  in 
the  construction  of  buildings  and  bridges  and  almost  entirely 
in  the  construction  of  the  hulls  of  modern  ships. 

Cast  steel,  as  well  as  cast  iron,  is  used  for  pistons  and  cross- 
heads  of  engines,  columns,  bed  plates,  bearing  pedestals  and  caps, 
propeller  blades,  and  for  many  small  pieces  and  fittings. 

Forged  steel  is  used  for  columns,  piston  rods,  connecting  rods, 
crank  and  line  shafting,  and  for  other  parts  of  engines  and 
machinery. 

84.  COPPER.  —  Copper  in  its  pure  state  is  red  in  color,  soft, 
ductile,  and  malleable,  with  a  melting  point  at  about  2000 
degrees,  and  a  tensile  strength  of  from  20,000  to  30,000  pounds 
per  square  inch  of  section.  It  is  not  readily  welded  except 
electrically,  but  is  easily  joined  by  the  operation  of  brazing. 
Attempts  have  been  made  to  temper  it,  but  without  practical 


200          THE  ELEMENTS  OF  MECHANICS   OF  MATERIALS 

success.  It  is  readily  forged  and  cast,  and  when  cold,  may  be 
rolled  into  sheets  or  drawn  into  wire.  When  in  sheets  or  in 
small  pieces  it  may  be  spun,  flanged,  and  worked  under  the 
hammer. 

The  tensile  strength  of  copper  rapidly  falls  off  as  the  tempera- 
ture rises  above  400  degrees,  so  that  from  800  to  900  degrees  its 
strength  is  only  about  one-half  that  at  ordinary  temperatures. 
This  peculiarity  of  copper  should  be  borne  in  mind  when  it  is 
used  in  places  where  the  temperature  is  liable  to  rise  to  these 
figures.  If  copper  is  raised  nearly  to  its  melting  point  in  con- 
tact with  air,  it  readily  unites  with  oxygen  and  loses  its  strength 
in  large  degree,  becoming,  when  cool,  crumbly  and  brittle.  Cop- 
per in  this  condition  is  said  to  have  been  burned.  The  possi- 
bility of  thus  injuring  the  tenacity  of  copper  is  of  the  highest 
importance  in  connection  with  the  use  of  brazed  joints  in  steam 
pipes. 

Copper  unalloyed  is  used  chiefly  for  pipes  and  fittings,  espe- 
cially for  junctions,  elbows,  bends,  etc.  For  large  sizes  of  pipes 
and  fittings,  the  copper  is  made  in  sheets,  bent  and  formed  to 
the  desired  shape,  and  brazed  at  the  seams;  for  small  sizes  the 
same  general  process  is  followed,  or  the  metal  is  drawn  from  the 
solid  and  bent  as  desired  after  the  drawing. 

Copper  is  also  used  as  the  chief  ingredient  of  the  various 
brasses  and  bronzes. 

85.  LEAD.  —  Lead  is  a  very  soft,  dense  metal,  grayish  in 
color  after  exposure  to  the  air,  but  of  a  bright  silvery  luster 
when  freshly  cut.  Commercial  lead  often  contains  small  amounts 
of  iron,  copper,  silver,  and  antimony,  and  when  so  combined  is 
harder  than  the  pure  metal.  It  is  very  malleable  and  plastic. 
In  engineering,  lead  is  chiefly  of  value  as  an  ingredient  of  bear- 
ing metals  and  other  special  alloys.  Lead  piping  is  also  used  to 
some  extent  as  suction  and  delivery  pipes  for  water  where  the 
pressure  is  only  moderate,  and  where  the  readiness  with  which 


ENGINEERING  MATERIALS  2OI 

it   may  be   bent  and  fitted  adapts  it  for  use  in  contracted 
places. 

86.  TIN.  —  Tin  is  a  soft,  white,  lustrous  metal  with  great 
malleability.     Commercial  tin  usually  contains  small  portions 
of  many  other  substances,  such  as  lead,  iron,  copper,  arsenic, 
antimony,  and  bismuth.     It  is  largely  used  as  an  alloy  in  the 
various  bronzes  and  other  special  metals.     Tin  resists  corrosion 
well,  and  in  consequence  is  frequently  used  as  a  coating  for  con- 
denser tubes.     It  is  also  used  for  coating  iron  plates,  the  product 
being  the  so-called  tin  plate  of  commerce.      It  melts  at  about 
450  degrees,  which  corresponds  to  a  steam  pressure  of  400  pounds 
per  square  inch,  approximately.     Due  to  this  low  melting  point, 
tin  is  often  used  in  the  composition  for  safety  plugs  in  boilers. 

87.  ZINC.  —  Zinc,  or  "  spelter,"  as  it  is  often  commercially 
called,  is  a  brittle  and  moderately  hard  metal  with  a  very  crys- 
talline fracture.     The  impurities  most  commonly  found  in  zinc 
are  iron,  lead,  and  arsenic.     It  is  used  chiefly  as  an  ingredient  of 
the  different  brass  and  bronze  alloys,  and  for  coating  iron  and 
steel  plates  and  rods.     The  process  of  applying  zinc  for  such  a 
coating  is  called  galvanizing,  and  the  product  is  "galvanized  " 
iron  or  steel.     Electricity  is  not  used  in  the  process;  the  articles, 
after  being  well  cleaned,  are  simply  dipped  into  a  tank  of  melted 
zinc  and  then  withdrawn. 

88.  ALLOYS.  —  A  mixture  of  two  or  more  metals  is  called  an 
alloy.     The  properties  of  an  alloy  are  often  surprisingly  different 
from  those  of  its  ingredients.     The  melting  point  is  sometimes 
lower  than  that  of  any  of  the  ingredients,  while  the  strength, 
elastic  limit,  and  hardness  are  often  higher  than  for  any  one  of 
them. 

Mixtures  of  copper  and  zinc  are  called  brass.  Mixtures  of 
copper  and  tin,  or  of  copper,  tin,  and  zinc,  with  sometimes  other 
substances  in  small  proportions,  form  gun  metals,  compositions, 
and  bronzes.  These  terms  are  rather  loosely  employed. 


202          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

Brass  and  compositions  are  used  for  piping  and  pipe  fittings; 
globe,  gate,  check,  and  safety  valves;  condenser  tubes  and  shells. 
The  bronzes  are  employed  for  many  of  the  uses  of  brass  where 
more  hardness,  strength,  and  rigidity  are  required.  They  are 
used  with  success  as  a  material  for  propeller  blades. 

89.  TIMBER.  —  Timber  is  not  extensively  used  in  modern  en- 
gineering construction.     The  advance  made  in  the  production  of 
steel,  whereby  its  homogeneity  is  assured  and  its  superior  strength 
unquestioned,  has  caused  it  to  be  substituted  for  wood  whenever 
it  is  possible  and  profitable  to  do  so. 

Generally  speaking,  the  heaviest  and  darkest  colored  timber 
is  the  strongest,  and,  in  all  cases,  the  strength  of  timber  is  great- 
est in  the  direction  of  its  grain. 

The  locality  from  which  timber  comes,  the  season  of  its  cutting, 
and  the  duration  of  its  seasoning,  are  factors  in  its  strength,  and 
the  uncertainty  arising  therefrom  makes  it  advisable  to  use  a 
factor  of  safety  of  not  less  than  10  in  calculations  relating  to  the 
dimensions  of  timber  to  bear  given  loads. 

90.  CONCRETE. — Concrete  is  a  mechanical  mixture  of  cement, 
sand,  and  broken  stone  or  gravel,  in  the  proportions,  usually,  of 
i  :  2  :  4  or   i  :  3  :  6.     It  is  largely  used  for  laying  foundations 
for  buildings  and  bridges  in  wet  ground,  and  for  breakwaters  and 
sea  walls.     After  laying,  it  soon  hardens  to  a  strong  mass  which 
is  little  permeable  to  water. 

Concrete  is  not  in  the  market  as  a  manufactured  product  but 
must  be  made  as  needed.  Whatever  the  proportions  used,  there 
is  the  utmost  necessity  for  thorough  mixing,  water  being  added 
as  may  be  necessary  to  secure  coherency  in  the  mixture.  The 
use  of  machines  designed  for  the  purpose  secures  a  more  perfect 
mixture  than  that  attained  by  the  hand  process. 

91.  REINFORCED  CONCRETE.  —  Beams  and  columns  of  re- 
inforced concrete  are  products  designed  to  avoid  the  uncer- 
tainty   concerning    the    protection    from    corrosion    and    fire 


ENGINEERING  MATERIALS  203 

that  attends  the  use  of  the  skeleton  steel  frame  in  building 
construction. 

There  can  be  no  question  as  to  the  appropriate  use  of  steel  for 
constructive  purposes  in  such  open  structures  as  bridges  and 
steamships,  but  in  cases  where  a  steel  skeleton  is  vested  with 
the  strength  of  a  structure,  and  is  subsequently  incased  in 
terra  cotta,  stone,  or  brick,  precluding  visual  inspection,  there 
is  serious  question  as  to  the  propriety  of  its  use. 

The  use  of  concrete  for  constructive  purposes  was  common 
among  the  ancients,  and  the  fact  that  in  some  ruins,  as  they 
stand  to-day,  the  concrete  parts  remain  —  the  stone  having 
long  since  disappeared  —  is  conclusive  evidence  of  its  durabil- 
ity. There  is  abundant  evidence  also  that  iron  embedded  in 
concrete  is  protected  from  the  corrosive  influence  of  moisture 
and  from  the  ravages  of  fire. 

It  has  been  demonstrated  experimentally  that  concrete  is 
strong  in  compression  but  quite  weak  in  tension.  In  the  case 
of  a  concrete  beam  supported  at  the  ends  and  loaded  at  the 
middle,  it  has  been  shown  that  the  upper  side,  which  is  in 
compression,  is  capable  of  supporting  ten  times  the  load 
which  would  cause  failure  at  the  lower  side,  which  is  in 
tension. 

Having  in  steel  a  material  of  very  high  tensile  strength,  and 
in  concrete  a  material  possessing  high  compressive  strength  as 
well  as  the  properties  of  durability  and  impermeability  to 
moisture,  the  problem  arose  of  effecting  their  combination  so 
as  to  produce  a  composite  material  having  the  desirable  qualities 
of  both.  The  solution  of  this  problem  was  the  production  of 
reinforced  concrete. 

In  the  case  of  beams,  steel  bars  are  embedded  in  the  area  of 
the  concrete  below  the  neutral  axis  to  reinforce  the  concrete 
subjected  to  tensile  stress,  thus  enabling  the  full  strength  of  the 
compression  area  above  the  neutral  axis  to  be  utilized  in  sup- 


204          THE  ELEMENTS  OF  MECHANICS  OF  MATERIALS 

porting  heavier  loads  than  would  have  been  possible  without  the 
combination. 

In  the  case  of  columns,  the  well-known  fact  that  the  most  eco- 
nomical metal  section  is  that  of  the  hollow  cylinder  suggested 
at  once  the  conception  of  a  column  having  three  concentric 
parts,  viz.,  a  central  core  of  concrete,  an  intermediate  zone  of 
steel,  and  an  outer  zone  of  concrete.  The  steel  is  thus  protected 
from  fire  and  moisture  and  is  best  disposed  for  the  utilization  of 
its  maximum  strength. 

Between  the  tubular  reinforced  concrete  column  just  described 
and  the  plain  non-reinforced  concrete  column  there  are  a  variety 
of  possible  combinations.  In  the  form  generally  used,  applicable 
alike  to  columns  and  piles,  the  reinforcement  consists  of  vertical 
rods  of  steel  tied  together  by  a  system  of  horizontal  wires. 
These  horizontal  ties  not  only  keep  the  vertical  rods  in  position, 
but  materially  assist  in  preventing  flexure  in  them;  they  also 
prevent  lateral  bulging  of  the  concrete. 

The  reinforcement  for  both  beams  and  columns  is  first  placed 
in  position,  after  which  they  are  enveloped  by  a  wooden  form, 
or  mold,  into  which  the  concrete  is  dumped,  and  rammed  when 
necessary.  After  a  period  of  thirty  hours  the  form  may  be 
removed  and  the  product  allowed  to  season  for  several  weeks 
before  being  subjected  to  its  load. 

92,  Adhesion  of  Concrete  to  Steel.  —  It  has  been  shown  by 
experiment  that  the  concrete  on  the  compression  side  of  rein- 
forced concrete  beams  may  be  subjected  without  rupture  to  a 
stress  twenty  times  that  which  would  cause  failure  in  a  tension 
test  in  the  concrete  alone.  It  has  also  been  shown  that  rein- 
forced concrete  acquires  a  power  to  resist  crushing  which  is 
greater  than  the  sum  of  the  resistances  of  the  two  materials 
taken  separately.  Such  remarkable  results  could  not  be  obtained 
were  it  not  for  the  adhesion  between  the  concrete  and  steel,  such 
adhesion  offering  resistance  to  sliding  between  the  two  surfaces 


ENGINEERING   MATERIALS  205 

and  facilitating  the  transference  of  the  forces  from  one  surface 
to  the  other. 

The  bond  between  the  steel  and  concrete  occasioned  by  ad- 
hesion alone  is  liable  to  be  destroyed  by  internal  stresses  due  to 
shocks  and  vibrations,  and  from  unequal  expansions  resulting 
from  thermal  changes,  and  it  is  with  the  idea  of  strengthen- 
ing the  adhesion  bond  by  mechanical  means  that  the  rein- 
forcing bars  are  usually  twisted  or  have  projections  on  their 
surfaces. 

93.  Proportion  of  Reinforcement.  —  To  secure  a  uniform  dis- 
tribution of  the  stress  in  a  reinforced  concrete  section,  the  rein- 
forcement should  consist  of  steel  bars  of  small  section  distributed 
so  that  each  shall  bear  its  allotted  part  of  the  stress.  The  pro- 
portion of  steel  to  concrete  depends  directly  upon  the  ratio  of 
the  coefficients  of  elasticity  of  the  two  materials.  The  value  of 
E  for  steel  is  30,000,000  pounds  per  square  inch,  implying  that 
a  force  of  one  pound  would  extend  or  compress  a  bar  of  steel 

i  square  inch  in  area  by  -  -  of  its  original  length.     The 

30,000,000 

value  of  E  for  concrete  may  be  taken  as  3,000,000  pounds  per 
square  inch,  so  that  for  an  equal  extension  of  the  two  materials 
the  steel  will  bear  ten  times  the  stress  that  can  be  borne  by  the 
concrete. 

Suppose  a  bar  of  steel  i  square  inch  in  section  area  to  be  sur- 
rounded by  a  ring  of  concrete  i  square  inch  in  area;  and  suppose 
further  that  the  steel  be  subjected  to  a  direct  pull  that  would 
occasion  in  the  concrete  the  safe  allowable  tension  stress  of  50 
pounds  per  square  inch.  The  elongation  in  the  concrete  would 

then  be  50  X =  0.0000167  inch.     Considering  the  bond 

3,000,000 

between  the  steel  and  concrete  to  be  perfect,  the  steel  would 
suffer  an  equal  elongation,  occasioning  a  stress  in  the  steel  of 

0.0000167  X  30,000,000  =  500  pounds  per  square  inch, 


206          THE   ELEMENTS   OF  MECHANICS   OF  MATERIALS 

a  result  only  one-thirtieth  of  the  safe  allowable  tension  stress 
for  steel;  consequently,  the  sectional  area  of  the  steel  may  be 
reduced  to  one-thirtieth  square  inch,  thus  raising  its  stress  to 
15,000  pounds  per  square  inch  without  causing  failure  in  the 
.concrete. 


CHAPTER  X 
TESTING   MATERIALS 

94.  Stress.  —  The  application  of  external  forces  to  a  piece  of 
material  tends  to  change  its  shape,  and  this  tendency  induces 
internal  forces,  known  as  stresses,  which  offer  resistance  to  the 
change.     These  stresses  may  be  of  three  kinds: 

1.  If  the  external  force  be  applied  at  right  angles  to  the 
section,  and  acts  away  from  it,  the  stress  is  one  of  tension,  or  a 
tensile  stress. 

2.  If  the  external  force  acts  toward  the  section,  the  stress  is 
one  of  compression,  or  a  compressive  stress. 

3.  If  the  external  force  acts  parallel  to  the  section,  the  stress 
is  one  of  shear,  or  a  shearing  stress. 

It  is  a  fundamental  assumption  that  these  direct  stresses  are 
uniformly  distributed  over  the  section,  so  that  if  W  denotes  the 
external  force  or  load,  A  the  area  of  section,  and  S  the  unit 
stress,  we  must  have,  in  the  absence  of  rupture,  W  =  AS.  W  is 
usually  expressed  in  pounds  and  A  in  square  inches,  so  that  we 

W 

shall  have  for  the  unit  stress  S  =  —  in  pounds  per  square  inch. 

A. 

95.  Strain.  —  A  piece  of  material  which  is  stressed  by  the 
application  of  external  force  undergoes  some  change  in  its  dimen- 
sions, either  lengthened  or  shortened,  and  the  amount  of  this 
distortion  is  known  as  the  strain  due  to  the  external  force,  or 
load. 

96.  Different  Kinds  of  Tests.  —  Materials  are  tested  for  ten- 
sion by  pulling  apart  a  test  piece  of  specified  dimensions;  for 
compression,  by  crushing  a  piece  of  definite  dimensions;  for 

207 


208          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

transverse  strength,  by  supporting  a  piece  at  two  points  and 
breaking  or  bending  it  in  a  testing  machine  by  applying  a  load 
at  an  intermediate  point ;  for  torsion,  by  twisting  apart  a  piece  in 
a  machine  designed  for  the  purpose;  for  direct  shearing,  by  break- 
ing a  riveted  or  pin-joint  connection  in  a  machine;  for  impact, 
or  shock,  by  letting  a  weight  drop  through  a  definite  height,  and, 
by  its  blow,  develop  suddenly  the  stress  in  the  material. 

97.  Ultimate   Strength.  —  The   ultimate   strength   of  a   test 
piece  is  the  load  required  to  produce  fracture,  reduced  to  a 
square  inch  of  original  section;  or,  in  other  words,  it  is  the  ulti- 
mate or  highest  load  divided  by  the  original  area.     Thus,  if  the 
area  of  the  cross  section  of  a  test  piece  is  0.42  square  inch,  and 
the   load  producing  fracture  is   28,400  pounds,   the   ultimate 

strength  is  — —  -  =  67,620  pounds  per  square  inch. 
0.42 

98.  Elastic  Limit.  —  The  elastic  limit  is  the  load  per  square 
inch  of  area  that  will  just  produce  a  permanent  set  in  the  mate- 
rial.    Thus,  in  a  tension  test,  if  the  cross-section  area  of  the  test 
piece  be  0.7  square  inch,  and  a  permanent  set  just  be  produced 
by  a  load  of  28,000  pounds,  the  elastic  limit  is  40,000  pounds 
per  square  inch. 

99.  Factor  of  Safety.  —  To  insure  safety  in  engineering  con- 
struction the  stresses  due  to  the  working  load  must  not  exceed 
the  elastic  limit  of  the  material  used,  and  to  insure  this  provi- 
sion it  is  customary  to  make  the  working  load  very  considerably 
less  than  the  load  necessary  to  produce  fracture.     The  ratio 
between  the  breaking  load  and  the  safe  allowable  load  is  the 
factor  of  safety;  or  it  is  the  quotient  obtained  by  dividing  the 
ultimate  strength  by  the  working  stress. 

100.  Elongation.  —  The  increase  in  length  of  a  test  piece, 
measured  just  before  rupture,  divided  by  the  original  length  of 
the  piece,  expresses  the  elongation. 

When  a  load  is  first  applied  to  a  test  piece  the  elongation 


TESTING   MATERIALS  2OQ 

is  nearly  uniformly  distributed  throughout  the  whole  length. 
This  continues  until  the  piece  begins  to  contract  in  area  near 
the  point  of  final  rupture,  and  nearly  all  the  subsequent  elon- 
gation is  restricted  to  the  immediate  vicinity  of  this  point. 

In  expressing  the  elongation  of  any  material,  the  length  of 
the  test  piece  must  be  stated.  If  the  length  of  the  test  piece 
is  8  inches,  and  an  extension  in  length  of  2  inches  is  noted  just 
before  rupture,  the  elongation  is  then  expressed  as  25  per  cent 
in  8  inches. 

101.  Reduction  in  Area.  —  The  reduction  in  area  is  found  by 
dividing  the  difference  between  the  original  and  final  section 
areas  at  the  point  of  rupture  by  the  original  area,  expressing  the 
fraction  in  per  cent. 

102.  Testing  Machines.  —  Machines  devised  to  determine  the 
physical  properties  of  materials  are  of  two  general  classes,  — 
hydraulic  and  screw  gear.     With  either  class  the  load  is  applied 
to  the  specimen  to  be  tested  in  a  manner  that  enables  it  to  be 
read  instantly  on  some  form  of  weighing  machine. 

The  weighing  system  employed  with  the  hydraulic  machine 
consists  usually  of  registering  the  applied  load  by  means  of  a 
gage  which  records  the  pressure  within  the  cylinder  of  an 
hydraulic  press.  Hydraulic  machines,  with  the  exception  of 
the  one  devised  by  Emery,  are  lacking  in  sensitiveness  and  ac- 
curacy and  are  not  extensively  used. 

Of  the  screw  gear  type  of  machines,  those  of  Riehle  and 
Olsen  are  very  generally  used.  With  either  machine  the  loads 
are  applied  from  some  external  source  and  are  transmitted  by 
means  of  spur  and  bevel  gears  to  upright  screws. 

The  Riehle  machine  consists  of  two  heads,  one  fixed  and  the 
other  movable.  The  upper  head  is  fixed  and  is  supported  by 
two  cast-iron  columns  which  rest  on  the  weighing  table  of 
the  machine.  The  weighing  table  rests  on  steel  knife-edges  in 
the  levers  of  a  compound  system,  the  last  lever  of  which  is  the 


210          THE  ELEMENTS  OF  MECHANICS  OF   MATERIALS 

weighing  beam  of  the  machine.  Two  upright  pulling  screws, 
which  turn  in  long  bearings  and  have  the  main  gears  keyed  to 
their  lower  ends,  pass  through  nuts  in  the  movable  lower  head 
and  reach  nearly  to  the  under  side  of  the  upper  head.  The 
screws  raise  or  lower  the  movable  head  according  to  the  direc- 
tion in  which  they  revolve. 

In  making  a  tensile  test,  the  specimen  is  gripped  at  its  end 
by  jaws  in  the  two  heads.  Power  is  applied  from  an  external 
source,  either  hand  or  motor,  in  a  direction  to  lower  the  movable 
head  by  the  screws,  thus  transmitting  a  pull  to  the  specimen, 
thence  to  the  upper  head,  and  then  to  the  weighing  table.  The 
pressure  on  the  weighing  table  is  transmitted  through  the  lever 
system  to  the  weighing  beam.  On  the  weighing  beam  is  a 
counterpoise  which  the  operator  moves  along  the  beam  to  main- 
tain the  balance  as  the  load  is  gradually  applied,  and  thus  has 
constantly  under  observation  the  magnitude  of  the  applied  load. 

In  making  a  compression  test,  a  cylindrical  cast  iron  block  is 
bolted  to  the  under  side  of  the  movable  head  and  the  specimen 
is  placed  between  that  and  a  similar  block  on  the  weighing  table. 
As  the  screws  lower  the  movable  head  the  pressure  is  brought 
to  bear  on  the  specimen,  thence  through  the  weighing  table  and 
lever  system  to  the  weighing  beam. 

In  making  a  transverse  or  bending  test,  the  specimen  is  laid 
on  two  supports  which  rest  on  the  weighing  table.  As  the 
movable  head  is  drawn  down  by  the  screws,  a  projection  on 
the  under  side  of  the  head  bears  on  the  middle  of  the  specimen 
and  the  pressure  is  transmitted  to  the  weighing  beam,  as  in  the 
other  tests.  The  deflections  of  the  specimen  may  be  measured 
to  o.ooi  of  an  inch  by  attaching  an  instrument  known  as  a 
deflectometer ,  or  transverse  indicator. 

103.  Forms  of  Test  Specimens.  —  Test  specimens  are  made 
from  coupons  cut  from  the  finished  product  of  the  material  to 
be  tested,  and  are  of  prescribed  form.  The  American  Society 


TESTING   MATERIALS 


211 


for  Testing  Materials  recommends  the  form  of  specimen  shown 
in  Fig.  112  for  tensile  tests  of  metal  plates,  inch  lengths  being 


Parallel  section 


not  less  than  9" 


-. —  akout  3~  J 


-about-18^- 


Fig.  112. 

marked  with  center  punch  on  the  length  of  parallel  section. 
Figure  113  shows  the  form  prescribed  by  the  Navy  Department, 


-about 


-  about  18— 
Fig.  113. 


The  round  form  of  test  piece  shown  in  Fig.  114  is  that  used 
for  testing  manufactured  products  other  than  plates,  such  as 
shafts,  axles,  and  beams. 


about  3 


-about  18- 


Fig.   114- 

Figures  115  and  116  illustrate  the  cold  bending  and  angle 
tests  for  wrought  iron  and  steel.  The  material  must  stand 
these  tests  without  sign  of  fracture. 


212  THE   ELEMENTS   OF  MECHANICS   OF   MATERIALS 

The  tensile  test  of  metals  is  the  simplest  and  most  important, 
as  it  determines  the  physical  properties  of  ultimate  strength, 
yield  point,  elastic  limit,  and  percentage  of  elongation. 


en- 


v      \  /      / 

^u 

Fig.  115.  Fig.  116. 

104.  Ductility.  —  Materials  such  as  wrought  iron,  mild  steel, 
copper,  and  other  metals  which  may  be  lengthened  by  the  appli- 
cation of  an  external  force,  with  a  corresponding  decrease  in 
thickness  or  in  diameter,  possess  the  property  of  ductility. 

105.  Plasticity.  —  If  in  the  process  of  stretching  a  ductile 
material  the  load  be  removed  and  none  of  the  strain  disappears, 
the  material  is  said  to  have  passed  beyond  the  ductile  and  to 
have  entered  the  plastic  stage.     The  plasticity  of  a  material  is 
determined  by  the  final  elongation  and  contraction  in  area  of 
the  test  piece.     In  structures  subjected  to  live  loads  and  shocks 
it  is  as  important  to  know  the  power  of  the  material  used  to 
resist  deformation  as  it  is  to  know  its  ultimate  strength,  and 
for  that  reason  specifications  for  iron  and  steel  usually  require 
a  certain  percentage  of  elongation  and  contraction  of  area  in  a 
stated  length  of  test  piece. 

1 06.  Stress-strain  Diagram.  —  If  in  testing  a  material  the 
gradually  applied  loads  be  plotted  as  ordinates  and  the  corre- 
sponding strains  as  abscissas,  the  resulting  curve  is  known  as  a 
stress-strain  diagram. 

For  tension  tests  of  wrought  iron  and  mild  steel  such  a  dia- 
gram will  take  the  form  shown  in  Fig.  117. 

The  load  being  gradually  increased,  it  will  be  found  that 
within  a  certain  limit  the  strains,  or  extensions,  will  be  directly 


TESTING   MATERIALS 


213 


proportional  to  the  augmentations  in  the  load,  and  that  if  the 
stress  be  relieved  the  test  piece  will  return  to  its  original  length. 
During  this  period  the  material  is  said  to  be  perfectly  elastic,  and 
will  be  so  represented  in  the  diagram  by  the  straight  line  OA. 
By  continuing  the  gradual  increase  in  the  load  a  point  will  be 
reached  where  the  proportionality  between  the  strain  and  the 
augmentations  in  the  load  ceases,  the  strain  increasing  much 
more  rapidly  than  the  load;  and  if  the  stress  be  relieved  the 


Max.  Stress 


Breaking- 
Stress 


Strain 
Fig.   117. 

piece  will  not  return  to  its  original  length,  but  will  acquire  a 
permanent  set.  The  load  at  which  this  occurs  is  known  as  the 
elastic  limit  of  the  material. 

A  further  increase  in  the  load  very  soon  develops  a  point 
where  the  extension  increases  very  rapidly,  —  as  much  as  from 
10  to  15  times  its  previous  amount,  — known  as  the  yield  point. 
This  rapid  increase  in  the  extension  usually  occasions  an  appar- 
ent reduction  in  the  stress,  as  shown  by  the  fall  in  curvature 
beyond  the  point  B.  For  commercial  purposes  the  yield  point 
and  the  elastic  limit  are  taken  as  the  same  point,  and  the  ordi- 


214          THE  ELEMENTS  OF  MECHANICS   OF   MATERIALS 

nate  at  B  would  represent,  to  the  scale  of  the  diagram,  this  limit 
in  pounds  per  square  inch  of  section  of  the  material. 

Passing  the  yield  point,  the  strains  increase  much  faster  than 
the  loads,  but  if  the  stress  in  the  material  be  relieved  a  careful 
measurement  will  show  the  disappearance  of  a  small  portion  of 
the  extension,  indicating  the  existence  still  of  some  elasticity 
and  that  the  specimen  is  passing  through  the  ductile  stage. 

At  about  the  time  the  maximum  stress  is  reached  at  C,  the 
material  appears  to  have  reached  the  plastic  state,  the  extension 
increasing,  in  time,  without  increase  in  load.  Up  to  this  point 
the  strain  has  been  evenly  distributed  throughout  the  length  of 
the  specimen,  but  here  occurs  an  extension  and  reduction  in 
section  purely  local,  immediately  followed  by  rupture  at  D. 

Within  the  elastic  limit  the  extension  of  the  specimen  probably 
would  not  exceed  o.ooi  of  its  length,  so  it  is  quite  impossible 
to  make  direct  measurements  of  the  extensions  corresponding  to 
the  augmentations  in  load.  Some  form  of  extensometer  is  used 
to  make  these  measurements. 

107.  Extensometer.  —  The  instrument  designed  to  measure 
accurately  the  minute  extensions  within  the  elastic  limit  of  a 
specimen  during  a  tensile  test  is  known  as  an  extensometer. 
There  are  various  forms  of  this  instrument,  but  the  type  known 
as  the  Riehle-  Yale  is  in  very  general  use  and  gives  dependable 
results. 

It  consists  essentially  of  two  clamps,  which  are  fastened  to 
the  test  specimen  by  set  screws.  The  parallelism  of  the  clamps 
is  secured  by  a  squaring  gage  bar  which  fits  neatly  in  guides  in 
the  clamps,  a  set  screw  arrangement  permitting  the  distance 
between  the  clamps  to  be  varied  according  to  the  distance  be- 
tween the  punch  marks  on  the  specimen  —  usually  the  standard 
distance  of  8  inches  —  in  order  that  the  set  screws  of  the  clamps 
may  fit  exactly  in  the  punch  marks.  The  upper  clamp  has  two 
projecting  arms,  180  degrees  apart,  through  which  pass  two  in- 


TESTING   MATERIALS  215 

sulated  bars  which  are  connected  with  the  lower  clamp  in  circuit 
with  a  battery  and  bell.  The  lower  clamp  has  projecting  arms 
corresponding  with  those  of  the  upper  clamp,  and  through  them 
pass  two  micrometer  screws,  each  having  a  vertical  fleet  of  one 
inch  and  reading  to  o.oooi  of  an  inch. 

The  instrument  being  attached  to  the  specimen  as  indicated, 
and  the  squaring  gage  bar  removed,  it  is  ready  for  use.  An 
increment  of  load  being  applied,  the  elongation  of  the  specimen 
is  measured  by  taking  the  reading,  first  of  one  of  the  micrometer 
screws  and  then  of  the  other,  by  running  them  up  until  the  con- 
tact of  the  point  of  the  screw  with  the  insulated  bar  completes 
the  circuit  and  causes  the  bell  to  ring.  Another  increment  of 
load  is  then  added  and  the  readings  taken  again,  and  so  on  until 
the  elastic  limit  is  reached,  the  average  readings  of  the  two 
micrometer  screws  for  each  increment  of  load  giving  the  actual 
elongation  of  the  specimen.  The  instrument  is  removed  from 
the  specimen  after  the  elastic  limit  is  passed. 

1 08.  Tensile  Test  of  Steel.  —  Figure  118  is  an  illustration  of 
a  tensile  test  of  steel,  the  scales  being  small  in  order  to  keep  the 
diagram  within  the  limits  of  the  page. 

The  original  dimensions  of  the  test  piece  were:  Length, 
8  inches;  diameter,  0.75  inch;  area  of  section,  0.4418  square 
inch. 

The  final  dimensions  were:  Length,  10.2  inches;  diameter 
0.4843  inch;  area  of  section,  0.1842  square  inch. 

An  inspection  of  the  diagram  shows  the  elastic  limit  to  have 
been  reached  at  about  21,000  pounds. 

-m      ±-     v      '.L     r  •  21,000 

Elastic  limit  of  specimen  =  -     —  =47,530  pounds  per  square 

0.44^8 

inch. 

TT1,.  Maximum  load      20,700       ., 

Ultimate  stress  =     .  .  .  -  =  -^u —  =  67,220    pounds 

Original  area        0.4418 

per  square  inch. 


2l6          THE  ELEMENTS  OF  MECHANICS   OF   MATERIALS 


.~  .       .0-i-  (lO.2  —  8)  100 

Extension  in  8  inches  =  -  -  —  -  -  =  27.5  per  cent. 

o 

~  r  (0.4418  —  0.1842)  100 

Contraction  of  area  =  —  -=58.31  per  cent 


30000 


1"  2" 

Scales: -Loads,  0.2=2000  Ibs.j  Extensions,  full  size. 
Fig.  118. 


Loads. 

Extensions. 

Remarks. 

2,OOO 

0.0012 

4,000 

0.0024 

6,000 

o  .  0034 

8,000 

0.0047 

10,000 

o  .  0060 

12,000 

O.OO72 

14,000 

o  .  0083 

l6,000 

0.0095 

l8,000 

0.0106 

20,OOO 

0.0118 

22,000 

O.0220 

Yield  point 

26,OOO 

o  .  5300 

29,680 

I  .92OO 

Maximum  stress 

25,82O 

2.2000 

Breaking  stress 

TESTING  MATERIALS 


217 


The  load  was  gradually  applied  with  the  uniform  augmenta- 
tion of  2000  pounds,  and  the  data  of  the  test  was  tabulated  as 
shown  in  the  table. 

To  find  the  modulus  of  elasticity  we  proceed  as  follows : 
The  sum  of  the  extensions  up  to  and  including  the  18,000 
pound  load  —  a  point  well  within  the  limit  of  elasticity  —  is 
0.0533  inch,  and  the  sum  of  the  loads  is  90,000  pounds.     The 

mean  extension  for  2000  pounds  is,  therefore,  -'  ^    •  _  OOOI2 

45 
inch. 

SL 

By  Art.  31,  p.  63,  we  have  E  =  — >  in  which  L  is  the  original 

length  and  S  the  load  producing  the  extension  y.     Here  L  =  8 
inches,  S  =  2000  pounds,  and  y  =  0.0012  inch. 

8  X  2000 


Hence,      E  = 


0.4418  X  0.0012 


=  30,180,000  Ibs.  per  sq.  inch. 


109.  Compression  Tests.  —  In  testing  materials  for  compres- 
sion, the  specimens  are  not  longer  than  1.5  to  3  times  the  diam- 
eter. If  the  specimens  are  long,  the  failure  under  compression 
will  be  by  buckling  or  bending,  and  for  intermediate  lengths 
partly  by  crushing  and  partly  by  bending. 

AVERAGE  PHYSICAL  PROPERTIES  OF  MATERIALS. 


Material. 

Pounds  per  square  inch. 

Pounds. 

Elastic  limit. 

Ultimate  strength. 

Modulus 
of  elas- 
ticity. 

Weight 
per 
cubic 
foot. 

Ten- 
sion. 

Com- 
pres- 
sion. 

Shear- 
ing. 

Ten- 
sion. 

Com- 
pres- 
sion. 

Shear- 
ing. 

Hard  steel 

60,000 
36,000 
25,000 
90,000 
170,000 
28,000 
6,000 
7,000 

3,ooo 

33,000 

20,000 
24,000 
1,000 

24,000 
25,000 

100,000 

60,000 
50,000 

120,000 

170,000 
50,000 
25,000 
30,000 

300 

10,000 

60,000 

50,000 

31,000,000 
30,000,000 
27,000,000 
31,000,000 
33,000,000 
25,000,000 
15,000,000 
15,000,000 

3,000,000 

1,500,000 

490 
490 
490 
490 
490 
480 
450 
550 
ISO 
40 

Mild  steel  
Cast  steel  
Tool  steel,  unhardened 
Tool  steel,  hardened.  . 
Wrought  iron  
Cast  iron 

55.000 
90,000 
49,ooo 
2,500 
8,000 

40,000 
20,000 

Copper  

Concrete  
Timber  

1,400 
600 

PART    II 

THE    ELEMENTS    OF    POWER 
TRANSMISSION 


CHAPTER  I 
TRANSMISSION    OF    POWER    BY    BELTS    AND    ROPES 

1.  Flat  Belt  Gearing.  —  The  transmission  of  power  by  means 
of  belts  running  over  pulleys  is  an  important  and  familiar  me- 
chanical contrivance.     It  is  practically  noiseless,  and  may  be 
used  for  transmitting  power  through  a  distance  as  great  as 
30   feet   without   intervening   support.     For   greater   distances 
idle  or  binder  pulleys  are  generally  used  to  tighten  the  slack 
side  of  the  belt. 

The  principal  disadvantage  of  belt  drives  is  that  due  to  the 
slip  occasioned  by  the  freedom  of  the  belt  to  slip  over  the  pulley, 
rendering  the  transmission  not  so  positive  as  that  through  the 
medium  of  gear  wheels;  but  this  disadvantage  becomes  an 
advantage  in  preventing  shocks  in  cases  where  mechanisms 
at  rest  are  suddenly  thrown  into  gear. 

2.  Materials  for  Flat  Belts.  —  The  most  common  material 
for  flat  belting  is  leather,  though  cotton  and  India  rubber  are 
not  infrequently  used.     The  best  leather  belting  is  made  of 
oxhide,  the  strips  of  the  hide  being  tapered  at  the  ends  and 
cemented   together   under  great  pressure,   and   then   laced   or 
riveted  with  copper  rivets  and  washers.     The  thickness  of  the 

single- ply  belt  varies  from  ^  to  tV  inch.     If   a  greater  thick- 

219 


220  THE  ELEMENTS  OF  POWER  TRANSMISSION 

ness  is  required,  two  or  three  strips  are  cemented  together 
under  pressure,  thus  producing  two-ply  and  three-ply  belting 
in  varying  thickness  up  to  J  inch.  The  average  weight  of 
leather  belting  is  0.036  pound  to  the  cubic  inch. 

Cotton  belts  are  usually  made  by  stitching  together  canvas 
or  ducking,  but  are  sometimes  woven  solid.  Such  belts  are 
cheaper  and  equally  as  strong  as  those  made  of  leather,  but  the 
coefficient  of  friction  is  rather  low  unless  the  material  is  properly 
sized.  The  weight  of  cotton  belting  varies  with  the  material, 
but  the  average  of  a  cubic  inch  may  be  taken  as  0.034  pound. 

Rubber  belting  is  produced  by  treating  canvas  with  a  com- 
position of  rubber  in  such  manner  as  to  fill  all  its  interstices.  It 
is  then  wrapped  in  rubber  and  vulcanized  under  heat  and  pres- 
sure. It  is  adaptable  to  use  in  damp  places,  since  the  material 
is  not  affected  by  moisture. 

The  weight  of  rubber  belting  is  about  0.044  pound  per  cubic 
inch. 

3.  Strength  of  Leather  Belting.  —  The  ultimate  strength  of 
leather  used  for  belting  varies  from  3600  to  5000  pounds  per 
square  inch  of  section   and  the  strength  of  the  laced  joint  may 
be  taken  as  one-third  that  of  the  solid  leather.     Taking  a  factor 
of  safety  of  5,  the  safe  working  stress  of  a  laced  joint  varies  from 
240  to  330  pounds  per  square  inch  of  section  for  single-ply  belting, 
and  from  500  to  600  pounds  per  square  inch  of  section  for  double- 
ply.     The  width  of  the  belt  must  be  made  sufficient  to  with- 
stand the  tension.     The  safe  tension  in  pounds  per  inch  of  width 
of  single-ply'  belting  ranges  from  50  to  80    according  to  the 
.thickness  and  the  safe  stress  of  section  of  the  material. 

4.  Coefficient  of  Friction  of  Leather  Belting.  —  The  coefficient 
of  friction  of  leather  belting  running  over  smooth  iron  pulleys 
is  a  very  important  but  exceedingly  variable  quantity,  ranging 
from   0.22    to   0.35.     With   wooden   pulleys   having  varnished 
faces  the  coefficient  of  friction  is  somewhat  greater. 


TRANSMISSION  BY   FLAT  BELTS  AND   ROPES  221 

5.  Velocity   of   Leather   Belting.  —  The   velocity   of   leather 
belting  varies  from  2000  to  6000  feet  per  minute,  but  the  most 
economical  speed  is  from  4000  to  5000  feet.     At  higher  speeds 
the  effect  of  centrifugal  action  is  excessive  and  the  life  of  the 
belt  shortened. 

6.  Velocity  Ratio  in  Flat-Belt  Transmission.  —  If  there  is  no 
slipping  of  the  belt  on  the  pulleys  when  transmitting  motion 
from  one  pulley  to  another,  the  outer  surface  of  the  rim  of  each 
pulley  will  have  the  same  velocity  as  the  belt;  so  if  we  denote 
by  DI  and  A  the  diameters  of  the  driver  and  follower  pul- 
leys respectively,  and  by  NI  and  N%  their  revolutions  per  minute, 
we  shall  have 

Speed  of  outer  surface  of  driver  rim      =  irDi  N\, 
and       Speed  of  outer  surface  of  follower  rim  =  irD^N^ 

and  since  each  of  these  is  equal  to  the  speed  of  the  belt,  we  have 
irAtfi  =  irD2NZl    whence    ^  =  ~' 

J\2          A 

That  is,   the  velocities  of  the  pulleys  are  inversely  as  their 
diameters. 


Fig.    i. 

When  motion  is  transmitted  from  one  shaft  to  another  by 
belting,  one  or  more  shafts  intervening,  as  in  Fig.  i,  we  shall  have 

J^l_A        ^3_A  AL5_A 

tf.'A'   #4     A'  tf.~A" 


222  THE  ELEMENTS  OF  POWER  TRANSMISSION 

Taking  the  product  of  these  equations,  member  by  member, 
and  remembering  that  N2  =  Na,  and  N^  =  N5,  we  have 

^i  =  A  v  A  v  A 

N6     A     A     A* 

That  is,  the  ratio  of  the  speed  of  the  first  pulley  to  the  speed 
of  the  last  pulley  is  equal  to  the  continued  product  of  the  ratios 
of  the  diameters  of  each  pair  of  pulleys  taken  in  order. 

The  results  just  obtained  are  true  only  for  very  thin  belts, 
or  for  belts  whose  thickness  is  so  small  in  comparison  with  the 
diameters  of  the  pulleys  as  to  be  negligible.  The  thickness  of 
belts  ordinarily  used  may  have  an  appreciable  effect  on  the 
velocity  ratio  of  two  pulleys.  While  the  belt  is  in  contact  with 
the  pulley,  its  inner  surface  is  in  compression  and  its  outer 
surface  in  tension,  but  the  neutral  surface  midway  between  is 
of  constant  length.  It  follows  that  the  velocity  of  the  surface 
of  the  belt  in  contact  with  the  pulley  is  less  than  the  velocity  of 
the  neutral  surface,  and  that  the  true  velocity  ratio  of  two 
pulleys  is  obtained  by  increasing  the  diameters  of  the  pulleys 
by  an  amount  equal  to  the  thickness  of  the  belt. 

7.  Slip  of  Flat  Belts.  —  Another  error  in  the  velocity  ratio 
of  belt-driven  pulleys  is  due  to  slip.  The  driving  side  of  a  belt 
is  necessarily  stretched  more  than  the  slack  side,  and  in  con- 
sequence the  driving  pulley  receives  a  greater  length  of  belt 
than  it  gives  off  to  the  follower  pulley;  therefore  the  speed  of 
the  driving  side  is  a  trifle  greater  than  that  of  the  slack  side. 
But  the  speed  of  the  rim  of  a  pulley  is  the  same  as  that  of  the 
belt  it  receives;  therefore  the  speed  of  the  rim  of  the  driver  will 
be  somewhat  greater  than  that  of  the  rim  of  the  follower,  and, 
as  a  consequence,  the  speed  of  the  follower  will  be  less  than  that 

given  by  the  ratio  equation  —  =  — ^-     This  difference  between 

NZ      A 

the  speeds  of  the  rims  of  the  driver  and  the  follower  is  known 
as  slip,  and  amounts  to  about  2  per  cent. 


TRANSMISSION  BY   FLAT   BELTS  AND   ROPES  223 

Example  I.  —  The  pulley  on  the  shaft  of  a  steam  engine  is 
4  feet  in  diameter,  from  which  a  belt  passes  to  a  pulley  2  feet 
in  diameter  on  a  shaft  in  a  room  above;  a  belt  passes  from  an- 
other 2-foot  pulley  on  this  shaft  to  one  of  10  inches  in  diameter 
on  a  third  shaft;  an  1 8-inch  pulley  on  the  third  shaft  is  belted 
to  a  6-inch  pulley  on  the  spindle  of  a  dynamo.  Find  the  speed 
of  the  dynamo  when  the  engine  is  making  160  revolutions  per 
minute. 

Solution.-         Speed  of  engine  =^ :  x  I?  x  1 JL, 
Speed  of  dynamo      48      24      18      72 
whence 

Speed  of  dynamo  =  -     — *—  =  2304  revolutions  per  minute. 

If  the  thickness  of  the  belts  in  this  example  were  0.25  inch  and 
the  slip  2  per  cent,  we  should  have 

Speed  of  dynamo  =  4  '2^  X  24       X  -— ^  X  160  X  0.98  =  2155 
24.25      10.25       6.25 

revolutions  per  minute,  showing  an  error  of  6.9  per  cent  in  com- 
puting the  speed  of  the  dynamo  by  the  usual  method. 

When  great  accuracy  is  required,  the  errors  due  to  slip  and 
to  the  failure  to  include  the  thickness  of  the  belt  in  the  pulley 
diameters  should  be  corrected. 

8.  Tensions  in  Flat  Belts.  —  When  a  belt  is  fitted  to  two 
pulleys  it  is  strained  over  them  while  at  rest  with  a  tension  TO, 
which  is  uniform  throughout  the  belt.  When  motion  ensues, 
the  driving  side  of  the  belt  stretches  and  its  tension  Ti  increases. 
At  the  same  time  the  slack  side  of  the  belt  is  shortened  and  its 
tension  T2  decreases.  The  theory  of  belting  rests  on  the  assump- 
tion of  perfect  elasticity  in  the  belts,  so  that  the  lengthening 
of  the  driving  side  of  the  belt  must  equal  in  amount  the  shorten- 
ing of  the  slack  side,  and  the  average  tension  in  the  belt  remains 
constant.  That  is,  we  should  have 

Ti  +  T2  =  2   TQ. 


224 


THE  ELEMENTS  OF  POWER  TRANSMISSION 


Such  is  not  the  case,  however,  as  the  materials  used  for  belting 
are  far  from  being  perfectly  elastic;  but  the  error  involved  is 
not  of  such  consequence  as  to  destroy  the  theory  when  applied 
to  the  conditions  under  which  belts  are  ordinarily  used. 

9.   Frictional  Resistance  between  a  Flat  Belt  and  a  Pulley.  - 
Let  the  two  pulleys  of  Fig.  2  be  joined  with  an  endless  belt. 
If  a  force  tends  to  turn  the  driving  pulley  D  in  the  direction 
indicated  by  the  arrow,  the  lower  part  of  the  belt  will  be  stretched 


Fig.   2. 


so  that  its  tension  T\  will  be  increased,  and  the  tension  T2  in 
the  upper  part  of  the  belt  will  be  decreased  an  equal  amount. 
Each  element  of  the  belt  in  contact  with  the  pulley  assists  the 
action  of  the  tension  in  the  slack  side  of  the  belt  in  resisting  the 
tension  in  the  tight  side,  and  therefore  the  tension  in  that  part 
of  the  belt  in  contact  with  the  pulley  varies  at  every  point. 
When  the  difference,  T\  —  T2,  becomes  sufficient  to  overcome 
the  resistance  to  motion  in  the  driven  pulley  F,  its  rotation 
begins.  The  difference  in  tensions  in  the  two  sides  of  the  belt 
is  the  amount  of  friction  between  the  belt  and  the  pulley,  and  is 
the  measure  of  the  driving  force.  If  the  tensions  were  equal, 
their  moments  about  the  center  of  the  driven  pulley  would  be 
equal  and  there  would  be  no  rotation,  since  there  would  be  equal 
turning  tendencies  in  opposite  directions. 

Let  TI  and  T2,  Fig.  3,  denote  respectively  the  tensions  in  the 
tight  and  slack  parts  of  the  belt  that  are  not  in  contact  with 


TRANSMISSION  BY  FLAT  BELTS  AND  ROPES 


225 


the  pulley.     Suppose  da  to  be  a  very  small  part  of  the  central 
angle  a  subtended  by  the  arc  of  contact  of  the  belt. 


Teos 


da. 


Fig.   3- 


Considering  the  equilibrium  of  the  very  small  arc  subtending 
da,  we  will  assume  the  tensions  at  its  extremities  to  be  T  and 
T  +  dT.  If  the  resulting  reaction  between  the  belt  and  the 
pulley  rim  due  to  these  tensions  be  denoted  by  R,  we  shall  have 
the  static  equation 

R  =  2  T  sin  —  =  Tda, 

2 

since  da  is  very  small  and  sin  —  and  —  are  very  approximately 

2  2 

the  same. 

As  slipping  is  about  to  take  place,  dT  is  the  measure  of  the 
friction  over  the  small  arc  considered. 

Then  dT  =  »R  = 

in  which  v  is  the  coefficient  of  friction. 


226  THE  ELEMENTS  OF  POWER  TRANSMISSION 

7  rrt 

For  the  small  arc  considered,  -=-  =  pda,  and  for  the  whole 

arc  of  contact  we  have 

~  dT  _    fc 

~T~  ~  I 

..      •*•  ^0 

from  which 

j1 

log  TI  —  log  T<L  =  jua,     or     log  —  =  ij.ay 

T1 

whence  —  =  e*"*, 

T2 

in  which  e  =  2.718,  the  base  of  the  Naperian  system  of  logarithms, 
and  a  is  expressed  in  circular  measure.     If  a  is  given  in  degrees 

it  may  be  converted  into  radians  by  multiplying  by  ~-  • 

1 80 

Hence,  it  is  seen  that  the  ratio  of  the  tensions  depends  only 
upon  the  coefficient  of  friction  and  the  angle  at  the  center  sub- 
tended by  the  arc  of  contact  of  the  belt,  and  is  independent  of 
the  diameter  of  the  pulley. 

A  convenient  expression  for  the  ratio  of  the  tensions  is 

TI 

loglQ—   =   JU«  lOglQ  2.7l8    =   O.4343  ILOL. 

The  result  just  obtained  is  applicable  to  a  rope  making  several 
turns  about  a  post,  and  explains  how  it  is  possible  for  one  man 
to  check  the  headway  of  a  ship  when  docking  by  taking  several 
turns  of  a  hawser  about  a  post  on  the  dock  and  pulling  the  slack 
end  with  a  comparatively  small  force. 

Example  II.  —  A  hawser  from  a  ship  makes  three  turns  about 
a  post  and  is  pulled  with  a  force  of  50  pounds.  The  coefficient 
of  friction  being  0.35,  what  force  is  exerted  in  checking  the 
headway  of  the  ship? 

Solution.  — 
Here  T2  =  50,     a  =  2  TT  radians,   and   M  =  0.35. 


TRANSMISSION  BY  FLAT  BELTS  AND  ROPES  227 

Then  log^-1  =  0.4343  X  0.35  X  6*  =  2.86638, 

^2 

whence 

T 

-^  =  735-!5>  and  T±  =  735.15  X  50  =  36,757  pounds. 
^2 

10.  Transmission  of  Power  by  Flat  Belts.  —  In  a  belt  con- 
nection between  two  pulleys  the  arc  of  contact  to  be  considered 
is  that  of  the  smaller  pulley,  and  if  the  connection  is  by  open 
belt  the  arc  of  contact  of  the  smaller  pulley  will  be  less  than 
1  80  degrees,  but  with  a  crossed  belt  the  arc  of  contact  will  be 
greater  than  180  degrees.  It  follows  that  with  the  same  con- 
ditions, a  greater  power  may  be  transmitted  with  the  crossed 
belt.  The  crossed  belt,  however,  is  subject  to  great  wear,  and 
generally  its  use  is  restricted  to  cases  where  it  is  desired  to  have 
the  pulleys  turn  in  opposite  directions. 

The  friction  between  a  belt  and  pulley  limits  the  power  that 
can  be  transmitted.  When  overloaded,  a  belt  will  slip  rather 
than  break;  therefore  at  the  point  of  slipping  the  driving  force 
is  TI  —  T2,  and  we  shall  have 


H.P.  transmitted  = 


33,000  550 

according  as  to  whether  the  velocity  is  expressed  in  feet  per 
minute  or  in  feet  per  second. 

The  ratio  of  the  tensions  is  given  by 

i      Ti 

log  —  =  0.4343  tux. 

-/2 

Example  III.  —  An  open  belt  3  inches  wide  connects  a  pulley 
5  feet  in  diameter  with  one  18  inches  in  diameter,  the  distance 
between  the  pulley  centers  being  12  feet.  The  larger  pulley 
makes  200  revolutions  per  minute,  the  coefficient  of  friction  is 
0.35,  and  the  greatest  tension  in  the  belt  must  not  exceed  80 
pounds  per  inch  of  width.  Find  the  horse  power  transmitted. 


228  THE  ELEMENTS  OF  POWER  TRANSMISSION 

Solution. — 

Velocity  of  belt  in  feet  per  second  =  5?r  X  2°°  - 


60 


52.36. 


=  80  X  3  =  240  pounds. 


Referring  to  Fig.  4, 


sin  0  = 


144 


Fig. 


0.1458,    whence  6  =  8°  23'. 


Arc  of  contact  of  smaller  pulley  =  [180°  —2  (8°  23')]— 

100 

=  2.85  radians. 
log—1  =  o.4343M«  =  0.4343  X  0.35  X  2.85  =  0.43320, 


therefore 


I  , 

—  !-  =  2.71,    whence 
T2         ' 


TI        240 

-  =  —  — 
2.71      2.71 


=  88.56  pounds. 


Then 


H.P.  = 


~  Tz)v  =  (240  -  88.56)  52.36  = 
55°  55° 


ii.  Centrifugal  Action  in  Belts.  —  When  belts  are  run  at 
high  speed  the  tensions  are  greater  than  those  due  to  the  power 
transmitted  on  account  of  the  centrifugal  action  in  that  part 
of  the  belt  in  contact  with  the  pulley.  This  centrifugal  action 
also  diminishes  the  normal  pressure  of  the  belt  on  the  pulley 
rim,  and  therefore  decreases  the  frictional  resistance.  For 


TRANSMISSION  BY  FLAT  BELTS  AND   ROPES 


229 


speeds  beyond  3500  feet  per  minute  the  stress  due  to  cen- 
trifugal action  increases  rapidly  and  should  be  taken  into 
account. 

Let  T  denote  the  tension  due  to  centrifugal  action  in  the  belt 
of  Fig.  5,  the  arc  of  contact  being  a. 


cZFSin(90-f+0) 
dF 


Fig.  5- 

Consider  the  very  small  arc  of  the  pulley  subtending  the  angle 
dd,  and  let  dF  denote  the  centrifugal  force  set  up  in  the  part 
of  the  belt  in  contact  with  this  arc. 

We  have  the  static  equation 

2  T  cos  (90°  -  -\  =  fdF  sin  (90°  -  -  +  A 

dF=^?~ 
r 


or 


230  THE  ELEMENTS  OF  POWER  TRANSMISSION 

If  Wi  denotes  the  weight  of  the  belt  in  pounds  per  linear  foot, 

then  dm  =  — -  •  rd8, 

g 

,„      WirdOv2      Wiv*d0 
and  dF  =  -         -  =  —     —  • 

gr  g 

Then 

„    .     a       W&2  ra        fa         \  ,         WiD* 
2  T  sin  -  =  — -  -  I    ™  i         "  i  j/i  - 


2  sin  - 

2 


,  rr       2  TTiP2       2fTFi    ^      Wv2 

whence  2  T  = —  = -  -  = > 

g  g        r       gr 

Wv2 


and  T  = 


2  gr 


in  which  W  =  2rW\  is  the  weight  of  a  portion  of  the  belt  whose 
length  equals  the  diameter  of  the  pulley;  v  the  velocity  in  feet 
per  second;  g  the  acceleration  of  gravity,  32.2  feet  per  second 
per  second;  and  r  the  radius  of  the  pulley  in  feet. 

Example  IV.  —  Find  the  width  and  thickness  of  belt  neces- 
sary to  transmit  10  horse  power  to  a  1 5-inch  pulley  so  that  the 
greatest  tension  may  not  exceed  60  pounds  per  inch  width  of 
belt  when  the  pulley  makes  1200  revolutions  per  minute,  the 
weight  of  the  belt  per  square  foot  being  1.5  pounds,  the  coefficient 
of  friction  0.25,  and  the  arc  of  contact  of  the  belt  165  degrees. 
The  weight  of  a  cubic  inch  of  leather  is  0.036  pound. 

Solution:  — 

TT  X  15  X  1200  r 

v  =  —  -  =  78.54  feet  per  second. 

12   X  OO 

/  rr\  rr*    \ 

H.P.  per  second  = 


55° 


TRANSMISSION  BY  FLAT   BELTS  AND   ROPES  231 

whence  TI  —  T2  =  ^— ; =  70  pounds. 

78-54 

a  =  arc  of  contact  =  ^-^— — -  =  2.88  radians. 
1 80 

T 
We  have,    log-1  =  0.4343  /«*  =  0.4343  X  0.25  X  2.88  =  0.31270, 

whence  ~ •  =  2.05. 

L  2 

Then      2.05  T2  —  T2  =  70,    whence    T2  =  66.67  pounds, 

and  TI  =  2.05  X  66.67  =  136.67  pounds. 

Wv2 

The  effect  of  centrifugal  action  increases  the  tension  by 

2gr 

pounds.  If  w  denotes  the  width  of  the  belt  in  inches,  then 
W  =  weight  of  a  portion  of  the  belt  whose  length  equals  the 

diameter  of  the  pulley  =  2  rW\  =  weight  of  -      -   square  feet  of 

the  belting  =  ^-^ ~  pounds. 

12 

„,,  Wv2      2rwX  1.5  X  (78.54)2 

Then  ^r=  2  grX  i/  =23-95«>, 

and  TI  =  136.67  +  23.95  w- 

But  from  the  conditions  of  the  problem  the  tension  must 
not  exceed  60  w\ 
hence     60  w  =  136.67  +  23.95^,    whence   w  =  3. 79  inches. 

If  /  denotes  the  thickness  of  the  belt  in  inches,  then  144  t  is 
the  volume  in  cubic  inches  of  i  square  foot  of  the  belting,  and 
since  i  cubic  inch  of  leather  weighs  0.036  pound,  we  have 

144  /  X  0.036  =  1.5,  whence   /  =  0.29  inch. 

12.   Flat  Belt  Connections  between  Non-parallel  Shafts.  - 

Two  non-parallel  and  non-intersecting  shafts  may  be  connected 
by  a  belt,  provided  the  pulleys  are  so  placed  that  the  point  at 
which  the  belt  leaves  either  pulley  lies  in  the  plane  of  the  other 


232 


THE  ELEMENTS   OF  POWER  TRANSMISSION 


pulley,  and  provided  further  that  the  belt  runs  only  in  one 
direction. 

As  an  illustration,  Fig.  6  shows  the  arrangement  of  a  belt 
making  a  quarter  turn,  the  shafts  being  at  right  angles.     It  will 


Fig.  6. 

be  noted  that  the  belt  in  passing  from  pulley  A  approaches 
pulley  B  in  a  direction  at  right  angles  to  the  axis  of  pulley  B, 
and  in  passing  from  pulley  B  it  approaches  pulley  A  in  a  direction 
at  right  angles  to  the  axis  of  pulley  A.  It  will  be  noted,  too, 
that  a  plane  that  is  tangent  to  the  rim  of  one  pulley  and  per- 
pendicular to  the  axis  of  the  other  pulley  cuts  the  other  pulley 
in  the  middle  of  its  rim. 

Non-parallel  shafts,  whether  they  intersect  or  not,  may  be 
connected  by  a  belt  to  run  in  either  direction  by  means  of  inter- 


TRANSMISSION  BY  FLAT  BELTS  AND   ROPES  233 

mediate  guide  pulleys  placed  on  a  spindle  whose  axis  is  the 
intersection  of  the  middle  planes  of  the  principal  pulleys. 

13.  Rope    Gearing.  —  Ropes    running    over    pulleys    having 
F-shaped  grooves  in  their  rims  are  used  in  preference  to  belts 
in  cases  where  much  power  is  to  be  transmitted.     The  hori- 
zontal distance  between  the  pulley  shafts  may  be  as  great  as 
90  feet,   the  ropes  between  the  pulleys  hanging  in  catenary 
curves.     The  materials  of  which  non-metallic  ropes  are  usually 
made  are  cotton  and  manila  hemp,  the  former  being  the  better 
on  account  of  its  greater  flexibility  and  higher  coefficient  of 
friction. 

14.  Systems  of  Rope  Gearing.  —  There  are  two  systems  of 
rope  transmission,  known  as  the  multiple  and  the  continuous. 

In  the  multiple  system  there  is  an  endless  rope  for  each  groove 
of  the  pulley  system,  whereas  in  the  continuous  system  there 
is  but  one  endless  rope  which,  when  it  leaves  the  first  groove  of 
the  driving  pulley,  enters  the  first  groove  of  the  follower  pulley. 
Leaving  the  first  groove  of  the  follower,  the  rope  enters  the  second 
groove  of  the  driver  and  leaves  that  groove  to  enter  the  second 
groove  of  the  follower,  thence  to  the  third  groove  of  the  driver, 
and  so  on  until  leaving  the  last  groove  of  the  follower,  when  it 
is  directed  by  a  guide  pulley  to  enter  the  first  groove  of  the 
driver. 

The  continuous  system  is  particularly  adaptable  to  cases 
where  the  distance  between  the  driving  and  the  driven  shafts  is 
short,  and  it  has  the  distinct  advantage  of  having  but  one  splice. 
A  marked  disadvantage  lies  in  the  fact  that  a  breakage  of  the 
rope  disables  the  whole  system,  which  would  not  be  occasioned 
by  the  breakage  of  any  one  of  the  ropes  of  a  multiple  system. 

The  ropes  most  commonly  used  vary  in  size  from  i  inch  to 
2  inches  in  diameter,  though  smaller  sizes  are  used  for  general 
transmission  in  manufacturing  establishments. 

Each  turn  in  the  coil  of  the  rope  in  the  continuous  system 


234  THE  ELEMENTS   OF  POWER  TRANSMISSION 

has  the  same  tension,  so  that  the  driving  force  is  the  same  as 
that  of  a  multiple  system  having  the  same  number  of  separate 
ropes  as  the  continuous  system  has  turns  in  its  coil. 

15.  Strength,   Weight,  and  Velocity   of  Rope  Belts. —  The 
breaking  strength  of  ropes  varies  from  7000  to  12,000  pounds 
per  square  inch  of  section,  but  to  insure  durability  it  is  good 
practice  to  limit  the  working  stress  of  a  rope  to  about  150  pounds 
per  square  inch,  indicating  a  factor  of  safety  of  60.     The  working 
stress  of  a  rope  may  be  taken  as  i2od2,  in  which  d  is  the  diameter 
of  the  rope  in  inches. 

The  weights  per  linear  foot  of  manila  and  cotton  ropes  are 
given  very  approximately  by  0.3 d2  and  o.2&d2  respectively, 
d  being  the  diameter  in  inches. 

The  velocity  of  transmission  rope  varies  from  3000  to  6000 
feet  per  minute,  the  most  efficient  speed  being  4700  feet. 

1 6.  Frictional  Resistance  between  Rope  and  Grooved  Pulley. 
—  The  F-shaped  grooves  in  the  rims  of  pulleys  used  for  non- 
metallic  rope  drives  are  so  dimensioned  that  the  rope  presses 
only  on  the  sides  of  the  groove  and  not  on  the  bottom,  thus 
securing  a  greater  resistance  to  slipping. 

Figure  7  represents  a  pulley  of  a  rope  drive  and  a  cross  section 
of  the  groove  and  rope.  Let  a  denote  the  arc  of  contact  of  the 
rope  and  2  6  the  angle  of  the  groove.  Let  R  denote  the  resultant 
central  force  due  to  the  normal  pressures  P  and  P  of  the  rope 
on  the  sides  of  the  groove.  Consider  the  small  length  of  arc 
of  the  rope  subtending  the  angle  da,  and  let  Q  denote  the  re- 
sisting pressure  at  each  side  of  the  groove  in  contact  with  the 
rope.  We  have  the  static  equation 

2  Q  sin  6  =  R  =  2  T  sin  —  =  Tda.     (See  Art.  9,  p.  224.) 

The  measure  of  the  friction  over  the  small  arc  considered  is 
dr  =  MX2<2  =  juX  Tda  X  cosec  0, 


TRANSMISSION   BY   FLAT   BELTS   AND    ROPES 


235 


whence 

Then 

whence 


7  rr\ 

—  =  M  cosec  Bda. 


n 

—  =  AC  cosec  B  I    da, 
J-  JQ 

T 

7~  =  ^  cosec  0, 


Fig.    7. 

in  which  6  is  half  the  angle  between  the  sides  of  the  groove. 
As  the  groove  angle  is  usually  about  45  degrees,  the  value  of  B 
is  22.5  degrees,  and  cosec  6  =  2.6. 


236  THE    ELEMENTS   OF  POWER   TRANSMISSION 

T 

Then  log-1  =  2.6^0:, 

J-z 

in  which  n  is  the  coefficient  of  friction  and  a  the  arc  of  contact 
of  the  smaller  pulley. 

Comparing  this  result  with  that  obtained  in  Art.  9,  p.  224,  we 
see  that  the  logarithm  of  the  tension  ratio  of  the  grooved  pulley 
is  2.6  times  as  great  as  that  obtained  for  the  flat  pulley. 

The  coefficient  of  friction  for  well  lubricated  manila  ropes  lies 
between  0.12  and  0.15,  and  for  cotton  ropes  between  0.18  and 
0.28. 

T 

The  ratio  — r  for  ropes  is  much  greater  than  that  for  flat  belts 
Tz 

owing  to  the  wedging  action  in  the  grooves. 

The  effect  of  centrifugal  action  in  ropes  is  appreciable  and 
may  be  determined  in  the  same  manner  as  for  flat  belts. 

17.  Transmission  of  Power  by  Ropes.  —  As  in  the  case  of 
flat  belts,  the  driving  force  in  rope  transmission  is  TI  —  T2,  and 
the  equation  for  the  tension  ratio  of  Art.  9,  p.  224,  is  applicable 
in  determining  the  power  transmitted  by  ropes  if  we  substitute 
IJL  cosec  6  for  ju,  0  being  half  the  angle  of  the  groove.  We  shall 
then  have 

(rp       y    \    -.  y 

H.P.  per  rope  =  —       — —  >    and  log  ~  =  na  cosec  0  =  2.6  pa, 

550  T2 

when  2  6  =  45  degrees,  which  is  commonly  the  case. 

Example.  —  How  many  ropes  of  1.125  inches  in  diameter  will 
be  required  to  transmit  200  H.P.  from  a  pulley  6  feet  in  diameter 
making  250  revolutions  per  minute,  the  coefficient  of  friction 
being  0.15  and  the  arc  of  contact  150  degrees? 
Solution.  — 

TI  =  i2od2  =  120  X  (i.i25)2  =  151.88  pounds. 

Arc  of  contact  =  -~—  =  2.618  radians. 
1 80 

T 

log—  =  2.6  jJLCt  =  2.6  X  0.15  X  2.6l8  =   I.02IO2 


TRANSMISSION    BY    FLAT   BELTS   AND   ROPES  237 

therefore         —-  =  10.5    and     T2  =  -^— 1  -  =  14.46  pounds. 
1 2  10.5 

\J    1        '4.  £  7T   X   6    X    250  - 

Velocity  of  ropes  =  -  — —  =  78.54  feet  per  second. 

60 

(Ti  -  T2)v      (151.88  -  14.46)78.54 
H.P.  per  rope  =  -  ^L— =  ^^—  JJ. — ^=10.62. 

55°  55° 

Number  of  ropes  required  =  -      -  =  10.2,  say  10. 

19.62 

18.  Telodynamic  Transmission.  —  The  method  of  trans- 
mitting power  over  long  distances  by  means  of  wire  ropes  and 
pulleys  has  been  given  the  name  of  telodynamic  transmission. 
Wire  ropes  are  enormously  stronger  than  ropes  made  of  manila 
and  cotton,  and  when  run  at  high  velocities  are  very  efficient 
in  the  transmission  of  power.  Their  excessive  wear  and  great 
cost  of  replacement  have  proven  to  be  such  serious  disadvantages 
as  to  restrict  the  employment  of  a  system  of  transmission  other- 
wise admirable. 

The  wire  ropes  commonly  used  have  six  strands,  each  of  which 
contains  six  wires  of  diameters  varying  from  0.02  to  0.083 
inch.  The  strands  are  wound  around  a  central  core  of  hemp, 
and  then  the  six  strands  are  twisted  around  the  central  core  of 
the  rope,  also  made  of  hemp. 

Wire  ropes  are  made  of  iron  and  steel,  the  latter  material 
being  the  better.  The  working  stress  of  wire  rope  has  been 
fixed  at  25,600  pounds  per  square  inch  of  section,  and  the  weight 
per  foot  run  may  be  taken  as  1.34^  pounds,  in  which  d  is  the 
diameter  of  the  rope  in  inches. 

Owing  to  the  excessive  wear  which  would  be  occasioned  by 
wedging  wire  ropes  in  the  pulley  grooves,  as  is  done  with  hemp 
and  cotton  ropes,  the  grooves  are  so  made  as  to  permit  the  ropes 
to  ride  on  the  bottom. 

The  equations  found  in  Arts.  9  and  n  are  applicable  to  wire 
ropes,  the  coefficient  of  friction  being  about  0.24. 


238  THE  ELEMENTS   OF  POWER  TRANSMISSION 

PROBLEMS 

1.  Two  pulleys,  15  inches  and  37.5  inches  in  diameter,  are  connected  by 
a  belt.     If  the  1 5-inch  pulley  makes  500  revolutions  in  10  minutes,  how 
many  turns  will  be  made  by  the  other  pulley  in  25  minutes?      Ans.    500. 

2.  Sketch  an  arrangement  of  four  pulleys  with  belts  for  driving  a  fan 
at  1500  revolutions  per  minute  from  a  shaft  making  200  revolutions  per 
minute,  giving  the  diameters  of  the  pulleys  to  be  used. 

3.  An  engine  shaft,  making  n  revolutions  per  minute,  carries  a  56-inch 
pulley  which  drives,  by  means  of  a  belt,  a  36-inch  pulley  on  the  line  shaft. 
The  line  shaft  carries  another  pulley,  42  inches  in  diameter,  which  is  belted 
to  a  24-inch  pulley  on  a  counter  shaft.     Another  pulley  on  the  counter 
shaft  is  48  inches  in  diameter  and  is  belted  to  a  14-inch  pulley  on  the  spindle 
of  a  dynamo.     Find  the  number  of  revolutions  made  in  a  minute  by  the 
dynamo  spindle.  Ans.    9.337*. 

4.  The  flywheel  of  an  engine  is  28  inches  in  diameter  and  is  belted  to 
a  pulley  20  inches  in  diameter  on  another  shaft.    A  20-inch  pulley  on  the 
second  shaft  is  belted  to  a  lo-inch  pulley  on  a  third  shaft,  which  carries 
an  1 8-inch  pulley,  which,  in  turn,  is  belted  to  a  6-inch  pulley  on  the  spindle 
of  a  dynamo.     Find  the  speed  of  the  dynamo  when  the  engine  is  making 
90  revolutions  per  minute.    If  the  belt  thickness  of  three-sixteenths  inch 
be  considered,  find  the  per  cent  of  loss  in  the  speed  of  the  dynamo. 

Ans.    756  revolutions;  3.18  per  cent. 

5.  The  shaft  of  a  high-speed  engine,  which  is  making  300  revolutions 
per  minute,  carries  a  14-inch  pulley,  over  which  passes  a  belt  to  a  2o-inch 
pulley  on  another  shaft.  A  lo-inch  pulley  on  this  shaft  drives  a  2o-inch 
pulley  on  a  third  shaft  carrying  a  6-inch  pulley  which  is  to  be  belted  to 
the  spindle  of  a  machine  so  that  the  revolutions  of  the  spindle  may  be 
45  per  minute.     Find  the  diameter  of  the  pulley  on  the  spindle. 

Ans.    14  inches. 

6.  A  weight  of  8000  pounds  is  suspended  from  one  end  of  a  rope.     How 
many  turns  of  the  rope  must  be  taken  around  a  circular  beam  fixed  horizon- 
tally in  order  that  a  man,  who  can  pull  with  a  force  of  250  pounds,  may 
keep  the  rope  from  slipping,  supposing  the  coefficient  of  friction  to  be  0.2? 

Ans.    2.75. 

7.  By  taking  3  turns  of  a  rope  about  a  post,  and  holding  back  with  a 
force  of  1 80  pounds,  a  man  just  keeps  the  rope  from  slipping.     The  coeffi- 
cient of  friction  being  0.2,  find  the  weight  supported  at  the  other  end  of  the 
rope.  Ans.   7810  pounds. 

8.  The  pulley  on  an  engine  shaft  is  5  feet  in  diameter  and  makes  100 
revolutions  per  minute.     The  motion  is  transmitted  from  this  pulley  to 
the  main  shaft  by  a  belt  running  on  a  pulley,  the  difference  in  tensions 


TRANSMISSION  BY  FLAT  BELTS  AND   ROPES  239 

between  the  tight  and  slack  sides  of  the  belt  being  115  pounds.  What  is 
the  work  done  per  minute  in  overcoming  the  resistance  to  motion  of  the 
main  shaft?  Ans.  180,714  foot  pounds. 

9.  A  belt  having  a  linear  velocity  of  350  feet  per  minute  transmits  5  H.P. 
to  a  pulley.     Find  the  tension  in  the  driving  side,  supposing  it  to  be  double 
that  in  the  slack  side.  Ans.  943  pounds. 

10.  A  pulley  3  feet  6  inches  in  diameter,  and  making  150  revolutions 
per  minute,  drives  by  means  of  a  belt  a  machine  which  absorbs  7  H.P. 
What  must  be  the  width  of  the  belt  so  that  its  greatest  tension  shall  be 
70  pounds  per  inch  of  width,  it  being  assumed  that  the  tension  in  the  driving 
side  is  twice  that  in  the  slack  side ?  Ans.  4  inches. 

11.  Find  the  horse  power  that  may  be  transmitted  by  a  belt  8  inches 
wide  and  passing  over  a  20-inch  pulley  on  the  shaft  of  an  engine  which 
makes  350  revolutions  per  minute.     The  angle  at  the  center  subtended  by 
the  arc  of  contact  of  the  belt  is  160  degrees,  and  the  coefficient  of  friction  is 
0.4.     The  tension  in  the  driving  side  may  be  taken  as  80  pounds  per  inch 
of  width,  and  the  stress  per  square  inch  of  belt  section  must  not  exceed  300 
pounds.   Find  also  the  thickness  of  the  belt.    Ans.  23. 94 H. P.;  0.27  inch. 

12.  A  belt  is  to  transmit  2  H.P.  from  a  pulley  12  inches  in  diameter  on 
a  shaft  making  160  revolutions  per  minute.     Find:  (i)  the  tensions  in  the 
driving  and  in  the  slack  sides  of  the  belt  when  the  arc  of  contact  is  180 
degrees  and  the  coefficient  of  friction  is  0.3.     (2)  The  width  of  the  belt 
when  the  thickness  is  0.25  inch,  and  the  safe  working  stress  320  pounds  per 
square  inch  of  belt  section. 

Ans.    215.15  pounds;  83.85  pounds;  2.69  inches. 

13.  What  horse  power  will  a  belt  8  inches  wide  transmit  over  an  1 8-inch 
pulley  making  300  revolutions  per  minute,  the  weight  of  i  square  foot 
of  the  belt  being  1.29  pounds,  the  coefficient  of  friction  0.3,  the  arc  of  con- 
tact of  the  belt  160  degrees,  and  the  tension  per  inch  of  width  of  the  belt 
not  to  exceed  80  pounds?     Take  the  weight  of  a  foot  of  belting  i  square 
inch  in  section  as  0.43  pound.  Ans.    15.8. 

14.  A  leather  belt  is  required  to  transmit  2  H.P.  from  a  shaft  running 
at  80  revolutions  per  minute  to  a  shaft  running  at  160  revolutions  per 
minute.     Find  the  stresses  in  the  belt,  assuming  that  the  smaller  pulley 
is  12  inches  in  diameter,  and  that  the  ratio  of  the  tensions  in  the  tight  and 
slack  sides  of  the  belt  is  2.25  :  i.     Find  also  the  width  of  belt,  taking  the 
working  stress  at  100  pounds  per  inch  of  width. 

Ans.    236.3  pounds;  105  pounds;  2.36  inches. 

15.  Find  the  width  of  belt  necessary  to  transmit  10  H.P.  to  a  pulley 
1 2  inches  in  diameter  so  that  the  greatest  tension  may  not  exceed  40  pounds 
per  inch  of  width  when  the  pulley  makes  1500  revolutions  per  minute.    The 


240  THE   ELEMENTS   OF  POWER   TRANSMISSION 

weight  of  the  belt  per  square  foot  is  1.5  pounds,  the  coefficient  of  friction 
0.25,  and  the  arc  of  contact  180  degrees.  The  weight  of  a  cubic  inch  of 
leather  may  be  taken  as  0.036  pound.  (The  effect  of  the  thickness  of  the 
belt  and  of  centrifugal  action  must  be  taken  into  consideration.) 

Ans.  8.4  inches. 

1 6.  An  8-inch  belt  traveling  over  a  3o-inch  pulley  making  174  revolu- 
tions per  minute  transmits  18  H.P.     The  ratio  of  the  tensions  in  the  tight 
and  slack  sides  of  the  belt  is  3.06  :  i,  the  arc  of  contact  160  degrees,  and  the 
maximum  tension  allowed  per  inch  width  of  belt  80  pounds.     Taking  the 
velocity  of  the  neutral  surface  of  the  belt  as  the  true  velocity,  it  is  required 
to  find:  (i)  The  coefficient  of  friction.     (2)  The  thickness  of  the  belt. 

Ans.   0.4;   0.25   inch. 

17.  A  driving  shaft,  making  100  revolutions  per  minute,  carries  a  pulley 
22  inches  in  diameter,  from  which  a  belt  communicates  motion  to  a  1 2-inch 
pulley  on  a  countershaft.     On  the  countershaft  is  also  a  cone  pulley  having 
steps  of  8,  6,  and  4  inches  in  diameter,  which  gives  motion  to  another  cone 
pulley,  of  equal  steps,  on  a  lathe  spindle.     Sketch  the  arrangement  in  side 
and  end  elevations,  and  find  the  greatest  and  least  speeds  at  which  the 
lathe  spindle  can  revolve.  Ans.    366.66;    91.66. 

18.  Determine  the  horse   power  that  may  be  transmitted   by  a  belt 
6  inches  wide  and  0.25  inch  thick  running  at  a  speed  of  60  feet  per  second. 
The  tension  in  the  slack  side  of  the  belt  is  0.45  of  that  in  the  tight  side,  and 
the  maximum  allowable  stress  per  square  inch  of  belt  section  is  280  pounds. 
Taking  the  weight  of  a  cubic  inch  of  leather  as  0.036  pound,  to  what  extent 
does  the  effect  of  centrifugal  action  reduce  the  power  transmitted? 

Ans.  20.87;  I7-2  Per  cent. 

19.  It  is  required  to  transmit  16  H.P.  from  a  pulley  20  inches  in  diameter 
by  means  of  a  belt  which  embraces  only  two-ninths  of  the  circumference 
of  the  pulley.     The  thickness  of  the  belt  is  three-eighths  of  an  inch,  the 
safe  working  stress  is  300  pounds  per  square  inch  of  belt  section,  and  the 
pulley  speed  is  1 20  revolutions  per  minute.     Find  the  tensions  in  the  two 
parts  of  the  belt,  and  the  width  of  belt  required. 

Ans.    2174.2  pounds;  1333.8  pounds;  19.32  inches. 

20.  What  horse  power  will  be  transmitted  from  a  zo-foot  pulley  by  12 
ropes  of  1.5  inch  diameter,  the  revolutions  being  140  per  minute,  the  co- 
efficient of  friction  0.14,  and  the  arc  of  contact  130  degrees?      Ans.  306. 

21.  A  drive  of  20  ropes  transmits  600  H.P.  from  a  lo-foot  pulley  mak- 
ing 100  revolutions  per  minute,  the  arc  of  contact  of  the  smaller  pulley 
being  160  degrees  and  the  coefficient  of  friction  0.12.     What  is  the  pull  in 
each  rope?     What  should  be  the  diameter  of  the  ropes  ? 

Ans.  364  pounds;  1.75  inches. 


CHAPTER   II 
TRANSMISSION  BY  TOOTHED  WHEELS 

19.  Toothed  Wheel  Gearing.  —  The  usual  arrangement  of 
toothed  wheels  in  train  for  the  transmission  of  power  is  to  have 
two  wheels  of  unequal  size  on  each  shaft  except  the  first  and 
last,  making  the  smaller  wheel  of  a  pair  on  one  shaft  gear  with 
the  larger  of  the  pair  on  the  next  shaft  in  the  series.  The  two 
wheels  of  unequal  size  on  a  shaft  is  the  mechanical  equivalent 
of  a  lever  with  unequal  arms,  and  therefore  modifies  the  power 
that  may  be  transmitted. 

The  circles  of  Fig.  8  represent  the  pitch  circles  of  two  toothed, 
or  spur,  wheels  in  gear.  The  pitch  circles  of  two  wheels  in  gear 
are  circles  which  appear  to  roll 
upon  each  other  and  which  pass, 
approximately,  through  the  mid- 
dle of  the  elevation  of  the  teeth. 
They  may  be  regarded  as  the 
outlines  of  two  discs  which  roll 
together  by  the  friction  at  the 

o  •/  . 

r  Fig.     8. 

circumferences. 

There  can  be  no  slipping  of  one  pitch  circle  over  the  other, 
owing  to  the  teeth;  therefore  the  same  length  of  circumference 
of  each  must  pass  over  the  point  of  contact  in  a  given  time. 
If  D  and  d  represent  the  diameters  of  the  large  and  small  wheels 
respectively,  and  N  and  n  the  number  of  their  revolutions  per 
unit  of  time,  then 

=  irdn,    or    DN  =  dn. 
241 


242  THE  ELEMENTS  OF  POWER  TRANSMISSION 

The  teeth  on  the  two  wheels  are  the  same  size,  and  their 
number  will  be  proportional  to  the  diameters  of  the  wheels, 
and  we  may  write  AN  —  Bny  in  which  A  and  B  are  the  numbers 
of  teeth  of  the  large  and  small  wheels  respectively. 

The  pitch  of  the  teeth  is  the  distance  measured  on  the  cir- 
cumference of  the  pitch  circle  between  the  centers  of  two  con- 
secutive teeth.  The  radius  of  the  pitch  circle  of  a  spur  wheel 
may  be  found  by  dividing  the  product  of  the  pitch  and  number 
of  teeth  by  2  TT. 

The  formulas  for  belt  pulleys  hold  for  toothed  wheels,  for  the 
belt  performs  the  same  office  as  the  teeth  —  it  causes  the  cir- 
cumference of  each  pulley  to  move  over  the  same  distance  in 
the  same  time. 

With  toothed  gearing  the  slipping  of  belt  gearing  is  avoided 
and  an  exact  velocity  ratio  may  be  maintained,  provided  the 
teeth  are  carefully  constructed  on  certain  geometrical  principles. 

20.  Arrangement  of  a  Train  of  Toothed  Wheels.  —  An  ar- 
rangement of  a  train  of  toothed  wheels  for  the  transmission  of 
power  is  shown  in  Fig.  9,  where  two  wheels  of  unequal  size  are 
placed  on  each  axis,  except  the  first  and  last,  and  where  the 
smaller  wheel  of  any  pair  gears  with  the  larger  wheel  of  the  next 
pair  in  the  train.  On  the  first  and  last  axis  there  is  but  one 
wheel. 


Fig.  9. 


In  this  arrangement  for  power  transmission  the  driving  wheels 
are  smaller  than  the  followers,  the  revolutions  of  the  successive 


TRANSMISSION  BY  TOOTHED   WHEELS  243 

axles  decreasing  in  number.  Should  a  high  velocity  ratio  be 
the  object  desired,  the  arrangement  would  be  the  reverse  of  that 
for  power  transmission,  the  driving  wheels  being  the  larger, 
and,  in  consequence,  the  revolutions  of  the  successive  axles 
would  increase  in  number. 

Suppose  the  power  to  be  applied  at  wheel  A.  Then  wheels 
A,  C,  and  E  will  be  the  drivers  and  wheels  B,  D,  and  F  the 
followers.  Let  the  letters  denoting  the  wheels  denote  also  the 
numbers  of  teeth  in  the  wheels  respectively.  Let  NI,  NZ,  Nz 
denote  the  number  of  revolutions  in  a  period  of  time  of  the 
drivers  A,  C,  E  respectively;  and  let  »i,  n2,  nz  be  like  repre- 
sentations of  the  followers  B,  Z>,  F. 

From  what  has  been  shown  we  shall  have 

ANi  =  Bni,     CN2  =Dn2,     and     ENS  =  Fn*. 
Multiplying  these  equations,  member  by  member,  we  have 
ANi  X  CN2  X  ENZ  =  Bni  X  Dn2  X  Fn3. 

Since  B  and  C  are  fixed  on  the  same  shaft,  we  have  N2  =  n\\ 
and  for  the  same  reason  N^  =  n2. 

Hence    ANi  XC  XE  =  B  X  D  XFn3,    whence  ^  *^*f  =  ^ 

-O  X  -L/  X  F       1\  i* 

That  is,  the  product  of  the  number  of  teeth  in  all  the  drivers, 
divided  by  the  product  of  the  number  of  teeth  in  all  the  fol- 
lowers, is  equal  to  the  ratio  of  the  number  of  revolutions  of  the 
last  wheel  to  the  number  of  revolutions  of  the  first  wheel.  This 
ratio  is  the  value  of  the  train,  and  is  denoted  by  e. 

If,  in  Fig.  9,  A  =  18,  C  =  20,  E  =  24,  B  =  36,  D  =  40, 
F  =  72,  and  ^Vi  =  120,  we  shall  have 

18  X  20  X  24        ns          , 
e  =  -  -  =  — -  i   whence  nz  =  10, 

36  X  40  X  72       120 

that  is,  the  last  wheel  will  make  10  revolutions  while  the  first 
wheel  is  making  1 20  revolutions. 


244  THE   ELEMENTS   OF   POWER   TRANSMISSION 

When  any  number  of  wheels  are  in  gear,  no  two  of  them  being 
on  the  same  axis,  as  in  Fig.  10,  the  combination  is  the  equivalent 
only  of  a  single  pair  of  wheels,  viz.,  the  first  wheel  and  the  last 
wheel;  the  intervening  wheels  simply  transfer  the  motion  and 
determine  the  direction  of  rotation  of  the  last  wheel.  If  the 
number  of  idler  wheels  intervening  between  the  first  and  last 
wheel  be  odd,  the  direction  of  rotation  of  the  first  and  last 
wheels  will  be  the  same;  if  even,  the  rotations  will  be  in  opposite 
directions. 


Fig.    10. 

21.  Driving  End  and  Load  End  of  Gearing.  —  Generally 
speaking,  the  part  of  a  system  of  gearing  to  which  the  motive 
power  is  applied  is  called  the  driving  end,  and  the  part  at  which 
the  resistance  is  overcome,  or  at  which  the  useful  work  is  done, 
is  called  the  load  end.  This  applies  to  the  operation  of  all  ma- 
chines, and  in  general  terms  we  have  these  relations: 

,T  ,     .,        ,.         Movement  of  driving  end 
Velocity  ratio  =  —7T~      — r~  > 

Movement  of  load  end 

Mechanical  advantage  = 


Driving  force 

._..    .      •     i    re  •  Useful  work  performed 

Mechanical  efficiency  =  — — — 

Total  work  expended 

22.  Toothed  Gearing  of  Screw-cutting  Lathes.  —  The  toothed 
gearing  of  a  screw-cutting  lathe  is  a  very  important  application 
of  wheels  in  train.  The  driving  wheel  of  the  gearing  is  either 
fast  on  the  lathe  spindle  or  derives  motion  by  means  of  inter- 


TRANSMISSION   BY   TOOTHED   WHEELS  245 

mediate  gearing.  In  either  case  the  revolutions  of  the  first 
driver  are  the  same  as  those  of  the  lathe  spindle.  The  inter- 
mediate gearing  affords  a  ready  means  of  throwing  the  gear 
wheels  out  of  action  when  the  lathe  is  to  be  run  at  high  speeds, 
as  for  polishing;  it  also  enables  the  direction  of  rotation  of  the 
lead  screw  of  the  lathe  to  be  changed  at  will.  If,  as  is  usually 
the  case,  the  lead  screw  is  right-handed,  the  screw  to  be  cut  will 
be  right-handed  or  left-handed,  according  as  the  direction  of 
rotation  of  the  lead  screw  is  the  same  as,  or  different  from,  that 
of  the  lathe  spindle. 

There  are  two  systems  of  lathe  gearing, — simple  and  compound. 

23.  Simple  Lathe  Gearing.  —  Fig.  n  is  a  representation  of 
simple  lathe  gearing,  no  two  wheels  being  on  the  same  axis. 
Any  wheels  intervening  between  the  driv- 
ing wheel  A  and  the  wheel  C  on  the  lead 
screw  of  the  lathe  have  no  influence  other 
than  to  convey  the  motion,  so  that  the 
only  wheels  to  be  considered  are  the  driver 
A  and  the  wheel  C  on  the  lead  screw. 
If  the  gearing  be  such  that  the  lathe  spindle 
and  the  lead  screw  make  the  same  number 
of  turns,  the  thread  cut  will  have  the  same 
pitch  as  the  thread  on  the  lead  screw. 

In  all  other  cases  the  pitch  of  the  thread  ,-,. 

Fig.  ii. 

to  be  cut  will   be   finer   or  coarser   than 

that  on  the  lead  screw  in  the  exact  proportion  that  the  revolu- 
tions of  the  lathe  spindle  in  a  unit  of  time  are  greater  or  less 
than  those  of  the  lead  screw. 

The  driving  wheel  A  is  fast  on  the  lathe  spindle,  or  so  con- 
nected with  it  as  to  make  the  same  number  of  revolutions  as 
the  spindle.  The  wheel  C  is  keyed  on  the  lead  screw  of  the 
lathe  and  the  intermediate  wheel  B  is  an  idler,  serving  the 
purpose  of  making  the  wheels  A  and  C  rotate  in  the  same 


246  THE   ELEMENTS  OF  POWER   TRANSMISSION 

direction  and  also  of  filling  in  the  space  between  A  and  C,  the 
axis  of  B  being  adjustable  in  a  slotted  arm. 

The  lead  screw  of  the  lathe  works  in  a  nut  on  the  lathe  carriage 
which  carries  the  cutting  tool,  so  that  for  each  revolution  of 
the  wheel  C  the  cutting  tool  advances  a  distance  equal  to  the 
pitch  of  the  lead  screw,  and  it  depends  entirely  upon  the  ratio 
between  the  numbers  of  teeth  of  wheels  C  and  A  as  to  the  number 
of  turns  the  piece  of  work  upon  which  the  thread  is  to  be  cut 
will  make  while  the  cutting  tool  is  moving  through  that  distance. 
From  this  we  shall  have 

Revolutions  of  lead  screw     _  Pitch  of  screw  to  be  cut 
Revolutions  of  lathe  spindle          Pitch  of  lead  screw 

The  problem  of  screw-cutting  consists,  then,  in  finding  a 
train  of  wheels  in  which  we  shall  have 

Pitch  of  screw  to  be  cut 

c  — • 

Pitch  of  lead  screw 

Example.  —  It  is  desired  to  cut  a  screw  of  10  threads  to  the 
inch  with  a  lathe  whose  lead  screw  has  a  pitch  of  J  inch.  Find 
a  suitable  train  of  wheels. 


Solution.  — 


=        =  -  =  — 


Giving  20  teeth  to  the  wheel  A  and  50  teeth  to  wheel  C  is  one 
of  a  number  of  solutions. 

24.  Compound  Lathe  Gearing.  —  Fig.  1 2  is  a  representation 
of  compound  lathe  gearing.  The  two  wheels,  B  and  C,  of 
different  size  on  the  axis  intervening  between  the  driver  A  and 
the  wheel  D  on  the  lead  screw  are  factors  in  the  value  of  the 
train.  From  Art.  20  we  have 

A  XC 

"  BXD' 


TRANSMISSION   BY   TOOTHED   WHEELS  247 

Example.  —  It  is  desired  to  cut  a  screw  of  16  threads  to  the 
inch  with  a  lathe  whose  lead  screw  has  a  pitch  of  ^  inch.  Find 
a  suitable  train  of  compound  gearing. 


Fig.  12. 

Solution.-  e  =  i  =  I  =  !Xi  =  2£x^  =  ^C 
i       8      2      4      40      96      J5  X  D 

Hence,  referring  to  Fig.  12,  giving  20  teeth  to  A,  and  24,  40, 
and  66  to  C,  £,  and  D,  respectively,  is  a  probable  solution. 

In  order  to  cut  a  left-hand  thread  the  lathe  spindle  and  the 
lead  screw  must  turn  in  opposite  directions.  This  may  be 
effected  by  interposing  an  idle  wheel  between  C  and  D. 

25.  Back  Gear  Lathe  Attachment.  —  The  back  gear  at- 
tachment to  a  lathe  is  a  mechanical  arrangement  to  increase 
the  power  of  the  lathe  at  the  expense  of  the  speed.  The  cone 
pulley  to  which  the  pinion  A  is  attached,  Fig.  13,  is  loose  on 
the  lathe  spindle.  The  spur  wheel  B  is  keyed  to  the  spindle. 
The  spur  wheel  C  and  pinion  Z),  carried  on  the  shaft  E,  form  the 
back  gear  and  can,  at  will,  be  thrown  into  or  out  of  gear  with 
A  and  B. 


248 


THE   ELEMENTS   OF   POWER   TRANSMISSION 


The  motion  of  the  cone  pulley  may  be  conveyed  to  the  lathe 
spindle  in  two  ways:  (a)  The  back  gear  being  disengaged,  as 
showTi  in  Fig.  13,  the  spur  wheel  B  is  made  to  engage  with  the 
speed  cone  by  means  of  a  bolt,  thereby  giving  to  the  lathe  spindle 
the  same  revolutions  that  are  made  by  the  cone  pulley,  (b) 
Disengaging  the  wheel  B  from  the  speed  cone  and  throwing  the 
back  gear  into  gear  with  wheels  A  and  B,  the  motion  of  the  cone 
pulley  is  then  transmitted  to  the  lathe  spindle  by  means  of 
the  train  ACDB,  making  the  entire  system  consist  of  a  train 
of  which  the  pulley  on  the  countershaft  is  the  first  driver  and 
the  spur  wheel  B  the  last  follower. 

c 


Fig.  13- 

26.   Transmission  of  Power  by  Toothed  Wheels.  —  In  the 

different  mechanisms  employing  a  train  of  toothed  wheels  for 
the  transmission  of  power,  either  the  movement  of  a  small  power 
through  a  comparatively  great  distance  is  utilized  in  overcoming 
a  much  greater  resistance  through  a  much  smaller  distance;  or, 
conversely,  the  movement  of  a  large  power  through  a  small  dis- 
tance is  utilized  in  making  a  smaller  resistance  move  through  a 
much  greater  distance.  In  the  two  cases  the  desired  results  were 
power  and  velocity  respectively,  the  underlying  mechanical 


TRANSMISSION   BY   TOOTHED   WHEELS 


249 


principle  being  that,  what  is  gained  in  power  is  lost  in  speed, 
and  conversely. 

The  power  may  be  applied  by  hand  to  the  end  of  a  lever,  or 
to  the  crank  pin  of  an  engine,  the  lever  or  crank  to  be  rigidly 
connected  to  the  first  axis  of  the  train. 


Fig.    14. 

Suppose  the  wheels  B  and  C  of  Fig.  14  to  be  fixed  on  the 
axis  c.  The  driving  power  P  is  applied  through  the  wheel  A 
tangentially  to  the  pitch  surface  of  the  teeth  in  contact  at  b, 
the  tendency  being  to  turn  the  wheels  B  and  C  in  a  contra- 
clockwise  direction  about  axis  c.  The  action  of  P  is  resisted 
by  the  reaction  R  of  the  load  applied  tangentially  to  the  pitch 
surfaces  of  the  teeth  of  the  wheels  C  and  D  in  contact  at  dt  with 
a  tendency  to  turn  the  wheels  B  and  C  in  a  clockwise  direction 
about  the  axis  c.  With  these  opposite  turning  tendencies  we 
shall  have,  when  motion  is  about  to  take  place, 

P  X  cb  =  R  X  cd, 
whence 

Radius  of  wheel  C     Number  of  teeth  in  wheel  C 


cd 
cb 


Radius  of  wheel  B      Number  of  teeth  in  wheel  B 


Example.  —  The  double  purchase  wheel  work  of  Fig.  15  repre- 
sents a  common  arrangement  applied  to  hoisting  machinery, 
such  as  cranes.  The  numbers  attached  to  the  spur  wheels  and 
pinions  indicate  the  number  of  teeth  they  contain.  The  length 


250 


THE  ELEMENTS  OF  POWER  TRANSMISSION 


of  the  lever  handles  being  18  inches,  the  radius  of  the  drum 
10  inches,  and  the  power  applied  to  each  of  the  handles  40  pounds, 
it  is  required  to  find:  (a)  The  weight  raised  at  the  drum;  (b)  the 
tangential  pressures  between  the  teeth  of  the  wheels;  (c)  the 
horse  power  transmitted,  supposing  the  handles  to  make  24 
turns  per  minute. 


Solution.  —  Since  the  radii  of  wheels  are  proportional  to  the 
number  of  teeth,  we  may  denote  by  15  x  and  120  3;  the  radii  of 
the  first  pinion  and  the  last  spur  wheel  of  the  train  respectively. 

Denoting  the  length  of  the  power  handles  and  the  radius  of 
the  drum  by  a  and  b  respectively,  we  have,  by  the  principle  of 
moments,  these  equations: 

P  X  a  =  &  X  150:  (i) 

Pi  X  80  =  #2  X  40  (2) 

P2  X  TOO  =  R3  X  20  (3) 

Pz  X  1200:  =  W  X  b  (4) 


TRANSMISSION   BY  TOOTHED   WHEELS  251 

Multiplying  these  equations,  member  by  member,  remember- 
ing  that  PI  =  RI,  P2  =  R^  and  Pz  =  Rs,  we  have 

PaXSoX  ioo  X  120  =  15  X  40  X  20  X  Wb, 
whence 

P_  _      15  X  40  X  20      Z>_        b  _  J^  y  10 i_ 

W      80  X  ioo  X  1 20    a  a      80      18      144 

and  W  =  11,520  pounds. 

In  the  above  calculations  the  effect  of  friction  and  of  the 
diameter  of  the  rope  have  been  neglected.  Friction  would 
likely  reduce  the  result  as  much  as  30  per  cent,  and  the  effective 
drum  radius  would  be  the  radius  of  the  drum  plus  the  radius 
of  the  rope. 

The  tangential  pressures  between  the  surfaces  of  the  teeth 
in  contact  may  be  found  by  means  of  equations  (i),  (2),  (3),  and 
(4)  if  the  pitch  of  the  teeth  be  known. 

Suppose  the  pitch  of  the  teeth  to  be  1.25  inches.     Then 

r>   j-        t  <c         •  •  T-25  X  15       75 

Radius  of  first  pinion  =  — —  — -  =  -^  • 

2  7T  8  7T 

From  equation  (i)  we  have 

80  X  18  =  RI  X  ~~  9    whence   RI  =  482.5  pounds. 

8  7T 

From  equation  (2)  we  have 

482.5  X  80  =  R2  X  40,    whence  RZ  =  965  pounds. 
From  equation  (3)  we  have 

965  X  ioo  =  R3  X  20,    whence  ^3  =  4825  pounds. 

These  results  indicate  that  the  teeth  of  the  wheels  should  be 
made  stronger  as  the  drum  shaft  is  approached.  This  could 
be  done  by  using  a  coarser  pitch  for  the  last  pair  of  wheels  in 
gear. 

If  Q  denotes  the  tangential  pressure  at  the  pitch  surface  of 


252  THE  ELEMENTS  OF  POWER  TRANSMISSION 

any  wheel  in  the  train,  and  V  the  velocity  in  feet  per  minute 
of  the  wheel  at  the  pitch  surface,  then 

V 


H.P.  transmitted  = 

33,000 

and  this  is  constant  for  any  stage  of  the  transmission. 

Thus,  if  the  lever  handles  make  24  turns  per  minute  we  shall 
have  at  the  lever  handle  axle 

V         80  X  2  TT  X  18  X  24 


=  0.540. 

X  12 

Since  the  value  of  the  train  is  sV,  the  last  wheel  will  make 
I  $  =  T3o  of  a  revolution  in  a  minute,  and  the  radius  of  the  last 

,     ,  .   1.25  X  120      75 
wheel  is  —  —        -  =  —  • 

2  7T  .  7T 

Hence  at  the  drum  axle 

H.P.  =  4825  X  2  *x  75X3  =         g- 

33,000  X  12  X  TT  X  10 

In  like  manner,  the  horse  power  transmitted  from  axle  A  to 
axle  B  may  be  shown  to  be  the  same,  thus: 

1  1\    X   24. 

Axle  A  makes  —  -  -*  =  4.5  revolutions  per  minute,  and  the 
80 


X   2 

- 

80 

radius  of  the  pinion  of  40  teeth  is-1-^  -  —  =  —  •     Then 

2  7T  7T 

TT  _      ,  .        „  065    X    2  7T   X    25    X  4-5 

H.P.  from  axle  A  to  axle  B  =  *-^  —*  -  ^  =  0.548. 

33,000  XvrX  12 

In  illustration  of  a  combination  of  belt  and  toothed  wheel 
gears  this  example  is  given: 

The  motion  of  an  engine  shaft  is  communicated  to  a  15  -inch 
pulley  on  a  shaft  A  by  means  of  an  open  belt  passing  from  a 
36-inch  flywheel.  The  motion  of  shaft  A  is  transmitted  to  a 
shaft  B  by  means  of  spur  wheels  of  J-inch  pitch  and  of  43  and 
24  teeth,  an  idle  wheel  intervening.  The  motion  of  shaft  B  is 
communicated  to  the  spindle  of  a  fan  by  means  of  an  open 


TRANSMISSION  BY  TOOTHED   WHEELS  253 

belt  passing  over  an  1  8-inch  pulley  on  shaft  B  to  a  1  2-inch 
pulley  on  the  fan  spindle.  The  engine  makes  250  revolutions 
per  minute  and  has  a  stroke  of  12  inches;  the  mean  pressure 
on  the  crank  pin  is  90  pounds,  and  the  ratio  of  the  tensions 
in  the  tight  and  slack  sides  of  the  flywheel  belt  is  2.25  :  i. 
Find:  (a)  The  number  of  revolutions  per  minute  of  the  fan; 
(b)  the  tangential  pressure  at  the  point  of  contact  of  the  teeth 
in  gear;  (c)  the  horse  power  at  two  or  more  stages  of  the  trans- 
mission; (d)  the  difference  of  the  tensions  in  the  tight  and  slack 
sides  of  the  fan  belt. 

Solution.  — 

Revolutions  of  fan  per  minute  =  25°  X  36  X  43  X  18  =  j6 

15  X  24  X  12 

Taking  moments  about  the  axis  of  the  engine  shaft,  Fig.  16, 
we  have 

18  Ti  =  go  X  6  +  18  T2,    whence    7\  -  T2  =  30. 
Then  2.25  T2  —  T%  =  30,    whence    T2  =  24, 

and  TI  =  30  +  24  =  54  pounds. 

Radius  of  spur  wheel  of  43  teeth  =  -  —  —  ^  =  5.95,  say  6  inches. 

8    X    2  7T 

Revolutions  of  shaft  A  =  —  -  —  =  600  per  minute. 

i5 

To  find  the  tangential  pressure,  P,  at  the  point  of  contact,  a, 
of  the  teeth  in  gear,  we  take  moments  about  the  axis  of  shaft  A, 
thus 


whence 


and  this  pressure  is  transmitted  without  change  to  the  point 
of  contact,  b,  the  intervening  spur  wheel  having  no  effect  other 


254 


THE  ELEMENTS  OF  POWER  TRANSMISSION 


than   to  cause  the  fan  to  turn  in  the  same  direction  as  the 
engine. 


Fig.   16. 

,       (.  PV  QO   X    12  7T   X    250 

H.P.  at  engine  shaft  = = *—  =  2.142. 

33,000          33,000X12 

,    ..  (Ti-  T2)V     30  X  15^X600 

H.P.  at  shaft  A  =  —  — =  —  -=2.142, 

33,000  33>°°°  X  12 

when  considering  the  15 -inch  pulley  on  shaft  A. 

The  horse  power  being  the  same  at  all  points  in  the  trans- 
mission, we  shall  have,  when  considering  the  1 8-inch  pulley  on 
shaft  B  and  the  tensions  in  the  fan  belt, 


whence 


HP   -  2  142  =  (7V  ~  ^0  18  TT  X  600  X  43 
33,000  X  12  X  24 

TI   —  Ti'  =  13.9  pounds. 


TRANSMISSION  BY  TOOTHED   WHEELS  255 

This  last  result  may  be  obtained  by  a  consideration  of  the 
equilibrium  of  the  whole  system,  thus: 

The  resistance  R  between  the  teeth  in  contact  at  a  maintains 
with  TI  and  T2  the  equilibrium  about  the  axis  of  shaft  A.  The 
contraclockwise  moment  of  P  at  a  is  balanced  by  the  clockwise 
moment  of  R  at  b,  while  P  at  b  maintains  with  TI  and  T2'  the 
equilibrium  about  the  axis  of  shaft  B. 

The  radius  of  the  wheel  of  24  teeth  on  shaft  B  being  — — 

8   X   27T 

=  3.34  inches,  we  shall  have,  by  moments  about  the  axis  of  shaft  B, 

9  TV  =  3.34  P  +  9  TV, 
or  9  (Ti  ~  TV)  =  3.34  X  37-5, 

whence  T\  —  7Y  =  13.9  pounds. 

as  found  above. 

These  calculations  neglect  the  losses  in  the  transmission  due 
to  belt  slipping  and  friction  between  the  teeth  and  at  the 
bearings,  the  aggregate  of  which  would  probably  exceed  30  per 

cent. 

PROBLEMS 

1.  A,  B,  and  C  are  three  parallel  spindles,  A  carrying  a  spur  wheel  of 
52  teeth  which  gears  with  one  of  19  teeth  on  B.    On  B  is  another  wheel  of 
8 1  teeth  gearing  with  one  of  21  teeth  on  C.    While  A  is  making  15  turns, 
how  many  will  C  make?     How  many  will  B  make? 

Ans.    158.33;    41-05. 

2.  If  the  change  wheels  of  a  lathe  have  18,  30,  40,  50,  and  88  teeth,  show 
an  arrangement  for  cutting  a  screw  of  1 1  threads  to  the  inch,  the  lead  screw 
having  a  pitch  of  one- third  inch. 

3.  With  a  lead  screw  of  one-half  inch  pitch,  and  change  wheels  of  20, 
24,  30,  40,  55,  60,  80,  and  100,  show  a  selection  of  wheels  to  cut  screws  of 
6,  n,  and  16  threads  to  the  inch. 

4.  The  back  gear  of  a  lathe  is  in  use.    Diameter  of  pulley  on  counter 
shaft,  4.25  inches;  diameter  of  pulley  of  lathe,  8.75  inches;  pinion  on  cone 
pulley  of  lathe,  18  teeth;  spur  wheel  on  back  shaft,  58  teeth;  pinion  on  back 
shaft,  18  teeth;  spur  wheel  on  lathe  spindle,  58  teeth.    How  many  revolu- 
tions per  minute  will  the  lathe  spindle  make  when  the  countershaft  is  mak- 
ing 150  revolutions  per  minute?  Ans.    7.017. 


256  THE  ELEMENTS  OF  POWER  TRANSMISSION 

5.  Two  parallel  shafts,  whose  axes  are  to  be,  as  nearly  as  possible, 
30  inches  apart,  are  to  be  connected  by  a  pair  of  spur  wheels,  so  that  while 
the  driver  runs  at  100  revolutions  per  minute  the  follower  is  required  to 
run  at  only  25  revolutions  per  minute.     Sketch  the  arrangement,  and  mark 
on  each  wheel  its  pitch  diameter  and  the  number  of  its  teeth,  the  pitch  of 
the  teeth  being  1.25  inches.     Determine  also  the  exact  distance  apart  of 
the  two  shafts.  Ans.  29.8295  inches. 

6.  Make  a  sketch  of  a  back  gear  of  a  lathe.     If  the  two  wheels  have  63 
teeth  each,  and  each  pinion  25  teeth,  find  the  reduction  in  the  velocity 
ratio  of  the  lathe  due  to  the  back  gear.  Ans.    i  :  6.35. 

7.  The  double  purchase  wheelwork  of  a  crane  consists  of  a  pinion  of 
1 6  teeth  on  the  handle  axle;  a  wheel  and  pinion  of  64  and  20  teeth  respec- 
tively on  the  first  intermediate  axle;  a  wheel  and  pinion  of  80  and  18  teeth 
respectively  on  the  second  intermediate  axle,  and  a  wheel  of  90  teeth  on  the 
drum  axle.     The  power  handles  are  20  inches  long,  the  radius  of  the  drum 
ii  inches,  the  diameter  of  the  rope  2  inches,  and  the  pitch  of  the  teeth  1.25 
inches.     Neglecting  friction,  it  is  required  to  find:  (i)  The  power  that  must 
be  applied  at  the  handles  to  raise  9600  pounds  at  the  drum;  (2)  the  tan- 
gential pressures  between  the  teeth  of  the  wheels  in  gear;  (3)   the  H.P. 
transmitted,  supposing  the  handles  to  make  20  turns  per  minute. 

Ans.     72  Ibs.;  452.4  Ibs.;  1447.7  Ibs.;  6434  Ibs.;  0.4569  H.P. 

8.  The  wheelwork  of  a  crane  consists  of  a  pinion  of  11  teeth  gearing 
with  a  wheel  of  92  teeth,  and  of  a  pinion  of  12  teeth  gearing  with  a  wheel 
of  72  teeth  on  the  drum  axle.     The  lever  handle  being  18  inches  long,  and 
the  radius  of  the  drum  9  inches,  it  is  required  to  find  the  ratio  of  the  power 
to  the  weight  raised.  Ans.     I  :  100  nearly. 

9.  In  the  example  of  a  combination  of  belt  and  toothed  wheel  gears 
given  at  the  end  of  Chapter  II,  it  is  required  to  find  the  horse  power  of 
the  transmission  at  shafts  A  and  B  by  considering  the  spur  wheel  of  43 
teeth  on  shaft  A  and  the  one  of  24  teeth  on  shaft  B. 


INDEX 


Adhesion  of  concrete  to  steel,  204. 
Advantage,  mechanical,  244. 
Alloys,  201. 
Angle  of  shear,  96. 

Of  tWiSt,  96. 

Applications  of  center  of  gravity,  9. 

Area,  reduction  in,  209. 

Arrangement  of  a  train  of  toothed  wheels,  242. 

Arm  of  force,  2. 

of  couple,  25. 

Assumptions  in  theory  of  beams,  67. 
Axis,  neutral,  64. 

Back  gear  lathe  attachment,  247. 
Beam,  cantilever,  25. 
Beam  design,  67. 
Beam,  simple,  26. 

Beam  resting  on  three  supports,  82. 
uniformly  loaded,  34. 

uniformly  loaded  over  part  of  its  length,  45,  48. 
with  concentrated  load  at  middle,  33. 
with  concentrated  moving  load,  55. 
with  overhanging  ends  and  uniformly  loaded,  42 . 
with  overhanging  ends,  uniform  and  concentrated  loads,  51 
with  two  equal  and  symmetrically  placed  loads,  38. 
Beams,  assumptions  in  theory  of,  67. 
continuous,  81. 
deflection  of,  73. 

distinction  between,  and  girders,  146. 
examples  in  deflection  of,  75-84. 
examples  in  resilience  of,  109-111. 
resilience  of,  109. 
standard  I,  68. 
strength  and  stiffness  of,  85. 
table  of  relative  strength  and  stiffness  of,  87. 
theory  of,  62. 

Belts,  centrifugal  action  in,  228. 
flat  (see  Flat  belts). 

257 


2  5  INDEX 

Belting,  leather,  coefficient  of  friction  of,  220. 
strength  of,  220. 
velocity  of,  221. 
Bending  moment,  25. 

combined  with  twisting,  100. 

diagrams,  28. 

first  derivative  of  is  the  shear,  37. 

influence  line  of,  58. 

Bending-moment  and  shear  diagrams  of  cantilevers,  39. 
Bending  moments,  general  case  of,  26. 
Bessemer  steel,  193. 
Bow's  system  of  lettering,  114. 
Box  girder,  68. 

Braced  cantilevers,  stresses  in,  159. 
Brass,  201. 

Brazing,  of  cast  iron,  190. 
Bronzes,  201. 

Cantilever,  25. 

Cantilever  beams,  bending-moment  and  shear  diagrams  of,  39,  133,  134. 

stresses  in  braced,  159. 
Carbon,  graphitic,  185. 

influence  of  in  cast  iron,  185. 
Castings,  inspection  of,  189. 
Cast  iron,  184. 

brazing  of,  190. 
gray,  185. 

influence  of  carbon  in,  185. 
influence  of  manganese  in,  188. 
influence  of  phosphorus  in,  188. 
influence  of  silicon  in,  187. 
influence  of  sulphur  in,  188. 
'    malleable,  190. 
mottled,  1 86. 
shrinkage  of,  189. 
strength  and  hardness  of,  189. 
uses  in  engineering,  189. 
white,  186. 
Center  of  gravity,  3. 

applications  of,  9. 

of  arc  of  circle,  5. 

of  a  cone,  12. 

of  a  plane  figure,  6. 

of  portion  of  polygon  or  sector  of  circle,  10. 

of  semicircular  arc,  13. 

of  a  system,  4. 


INDEX  259 


Center  of  gravity,  of  a  triangle,  9. 
Center  of  gyration  (see  Radius  of  gyration). 
Center  of  moments,  i. 
Centrifugal  action  in  belts,  228. 
Coefficient  of  elasticity,  62. 

friction  in  leather  belting,  220. 

rigidity,  63. 
Columns,  89. 

design  of,  91. 

Combined  torsion  and  bending,  100. 
Compositions,  201. 
Compression  tests,  217. 
Concrete,  202. 

reinforced,  202. 
Cone,  center  of  gravity  of,  12. 
Continuous  beams,  81. 
Contrary  flexure,  points  of,  74. 
Copper,  199. 
Couple,  25. 

arm  of,  25. 
Crane,  framed,  177. 
Crucible  steel,  193. 

Dangerous  section,  36. 
Deficient  frames,  179. 
Deflection  of  beams,  73. 

examples  in,  75-84. 

Deformation,  internal  work  due  to,  105. 
Design  of  columns,  91. 
Diagram,  force,  116. 

frame,  148. 

of  bending  moments,  28. 

reciprocal,  148. 

shear,  29. 

stress,  148. 

stress-strain,  212. 
Driving  end  of  gearing,  244. 
Ductility,  212. 

Efficiency,  mechanical,  244. 
Elastic  limit,  208,  213. 
Elasticity,  coefficient  of,  62. 

modulus  of,  62. 

transverse,  63. 
Elongation,  208. 
Engineering  materials,  184. 


260  INDEX 

Equilibrium,  2. 

Equivalent  twisting  moment,  100. 

Extensometer,  214. 

Factor  of  safety,  208. 
Ferromanganese,  188. 
Fink  truss,  165. 
Flanges  of  I  beams,  68. 
Flat  belt  gearing,  219. 

and  pulley,  frictional  resistance  between,  224. 
connections  between  non-parallel  shafts,  231. 
transmission  of  power  by,  227. 
transmission,  velocity  ratio  in,  221. 
Flat  belts,  materials  for,  219. 
slip  of,  222. 
tensions  in,  223. 

transmission  of  power  by,  219,  227. 
Force  action  at  a  joint,  147. 
Force  diagram,  116. 

pole  of,  119. 
vectors  of,  119. 
Forces,  supporting,  146,  148. 
Forms  of  test  specimens,  210. 
Frame,  145. 

diagram,  148. 

force  action  at  joint  of,  147. 
Framed  crane,  177. 

structures,  145. 
Frames,  deficient,  179. 

redundant,  179. 
Frictional  resistance  between  flat  belt  and  pulley,  224. 

rope  and  grooved  pulley,  234. 
Funicular  polygon,  117. 

a  bending-moment  diagram,  125. 

examples  in  its  application  to  beams,  126-143. 

illustrations  of,  120. 

Galvanizing,  201. 

Gearing,  compound  lathe,  246. 

driving  end  of,  244. 

flat  belt,  219. 

load  end  of,  244. 

rope,  233. 

simple  lathe,  245. 

toothed  wheel,  241. 
Girder,  box,  68. 


INDEX  261 

Girder,  Linville  or  N,  164. 

Warren,  162. 

Girders,  distinction  between  beams  and,  146. 
Graphic  statics,  114. 

Gravity,  center  of  (see  Center  of  gravity). 
Gun  metals,  201. 
Gyration,  center  of  (see  Radius  of  gyration). 

Hooke's  law,  63. 

I  beams,  standard,  68. 

flanges  and  web  of,  68. 

Illustrations  of  moments  of  inertia  and  radii  of  gyration,  19-22. 
Inertia,  moments  of  about  parallel  axes,  relation  between,  18. 
Inertia,  plane  moment  of,  17. 

circular  surface,  21. 

parallelogram,  20. 
Inertia,  polar  moment  of,  17. 

right  circular  cone,  22. 
Inflection,  points  of,  74. 
Influence  line  of  bending  moment,  58. 

shear,  60. 

Internal  work  due  to  deformation,  105. 
Iron,  cast  (see  Cast  iron). 
Iron,  wrought  (see  Wrought  iron). 

Lathe,  back  gear  attachment  of,  247. 
compound  gearing  of,  246. 
simple  gearing  of,  245. 

Lathes,  screw  cutting,  toothed  gearing  of,  244. 
Lead,  200. 

Leather  belting,  coefficient  of  friction  of,  220. 
strength  of,  220. 
velocity  of,  221. 
weight  of,  220. 

Lettering,  Bow's  system  of,  114. 
Limit,  elastic,  208,  213. 
Linville  or  N  girder,  164. 
Load  end  of  gearing,  244. 
Loads,  dead,  145. 
live,  145. 
moving,  55. 
on  structures,  145. 
suddenly  applied,  effect  of,  105. 

Machines,  testing,  209. 
Manganese,  influence  in  cast  iron,  188. 
steel,  196. 


262  INDEX 

Materials,  engineering,  184. 

average  physical  properties  of,  217, 
different  kinds  of  tests  of,  207. 
Materials  for  flat  belts,  219. 
Mechanical  advantage,  244. 
Mechanical  efficiency,  244. 
Modulus  of  elasticity,  62. 
resilience,  107. 
rigidity,  63. 
section,  66. 

polar,  98. 
Moment,  i. 

equivalent  twisting,  100. 
resisting,  65,  97. 
twisting,  95. 

Moment  of  inertia  or  second  moment  (see  Inertia). 
Moments,  bending  (see  Bending  moments), 
center  of,  i. 
clockwise,  2. 
contraclockwise,  2. 
Moving  loads,  55. 

N  or  Linville  girder,  164. 
Neutral  axis,  64. 

surface,  64. 
Nickel  steel,  199. 

Open-hearth  steel,  194. 

Permanent  set,  213. 

Phosphorus,  influence  of  in  cast  iron,  188. 

steel,  196. 
Plasticity,  212. 
Point,  yield,  213. 

Points  of  inflection  or  of  contrary  flexure,  74. 
Polar  moment  of  inertia,  17. 
Pole  of  force  diagram,  119. 
Polygon,  funicular,  117. 

examples  in  its  application,  120. 
Power,  transmission  of  by  flat  belts,  227. 

ropes,  236. 

shafts,  100. 

toothed  wheels,  248. 

Radius  of  gyration,  plane,  19. 

of  circular  surface,  21. 
of  parallelogram,  20. 


INDEX  263 

Radius  of  gyration,  polar,  of  right  circular  cone,  22. 

Rankine's  formula  for  columns,  89. 

Ratio,  velocity,  244. 

Reactions,  support,  26,  148,  149. 

Reciprocal  diagram,  148. 

drawing  of,  151. 

examples  in  drawing  of,  152-156. 
Reduction  in  area,  209. 
Redundant  frames,  179. 
Reinforced  concrete,  202. 

proportion  of  reinforcement  in,  205. 
Resilience,  105. 

modulus  of,  107. 
of  beams,  109. 
of  torsion,  112. 

Resisting  moment  for  bending,  65. 
torsion,  97. 

Riehle  testing  machine,  209. 
Riehle-Yale  extensometer,  214. 
Rigidity,  coefficient  of,  63. 
Rolling  loads,  55. 
Roof  truss  fixed  at  the  ends  and  with  wind  pressure,  167. 

one  end,  free  at  the  other,  wind  and  dead  loads,  173. 
Rope  and  grooved  pulley,  frictional  resistance  between,  234. 
Rope  belts,  strength,  weight  and  velocity  of,  234. 
Rope  gearing,  233. 

systems  of,  233. 
Ropes,  transmission  of  power  by,  236. 

Safety,  factor  of,  208. 

Screw  cutting  lathes,  compound  gearing  of,  246. 
simple  gearing  of,  245. 
toothed  gearing  of,  244. 
Second  moment,  18. 
Section,  dangerous,  36. 

method  of  determining  stresses,  157. 
plane  modulus  of,  66. 
polar  modulus  of,  97. 
Semicircular  arc,  center  of  gravity  of,  13. 
Set,  permanent,  213. 
Shafts,  95. 

transmission  of  power  by,  100. 
Shear,  29. 

angle  of,  96. 
diagrams,  29. 
influence  line  of,  60. 


264  INDEX 

Silicon,  influence  in  cast  iron,  187. 

steel,  196. 

Simple  lathe  gearing,  245. 
Slip  in  flat  belts,  222. 
Specimens,  forms  of  test,  210. 
Spelter,  201. 
Spiegeleisen,  188. 
Statics  (see  Graphic  statics). 
Steel,  192. 

adhesion  of  concrete  to,  204. 

Bessemer  process  of  making,  193. 

Bessemer  and  open-hearth  compared,  195. 

cement  or  blister,  193. 

crucible,  193. 

influence  of  manganese  in,  196. 

influence  of  phosphorus  in,  196. 

influence  of  silicon  in,  196. 

influence  of  sulphur  in,  195. 

nickel,  199. 

semi,  196. 

Siemens-Martin  or  open-hearth  process,  194. 

structural,  193. 

tempering  of,  198. 

tension  test  of,  216. 

tungsten,  199. 

uses  in  engineering,  199. 
Steels,  special,  198. 
Strain,  62,  207. 

Strength  and  stiffness  of  beams,  85. 
of  leather  belting,  220. 
of  rope  belts,  234. 
ultimate,  208. 
Stress,  62,  207. 

kinds  of,  207. 
or  reciprocal  diagram,  148. 
rule  for  determining  kind  of,  156. 
section  method  of  determining,  157. 
Stress-strain  diagram,  212. 
Stresses  in  braced  cantilevers,  159. 
Structural  steel,  193. 
Structures,  framed,  145. 
Strut,  145. 

Suddenly  applied  loads,  effect  of,  105. 
Sulphur,  influence  of  in  cast  iron,  188. 

steel,  195. 
Support  reactions,  26,  148,  149. 


INDEX  265 

Supporting  forces,  146,  148. 
Surface,  neutral,  64. 
System  of  lettering,  Bow's,  114. 
Systems  of  rope  gearing,  233. 

Telodynamic  transmission,  237. 
Tension  test  of  steel,  216. 
Tensions  in  flat  belts,  223. 
Test  specimens,  forms  of,  210. 

compression,  217. 
Tests,  different  kinds  of,  207. 
Testing  machines,  209. 
Tie,  145. 
Timber,  202. 
Tin,  201. 
Tin  plate,  201. 

Toothed  gearing  of  screw-cutting  lathes,  244. 
Toothed  wheel  gearing,  241. 
Toothed  wheels,  arrangement  of  train  of,  242. 
transmission  of  power  by,  248. 
value  of  train  of,  243. 
Torque,  95. 
Torsion,  95. 

combined  with  bending,  100. 
resilience  of,  112. 

Transmission  of  power  by  flat  belts,  227. 
ropes,  236. 
shafts,  100. 
toothed  wheels,  248. 
Transmission,  telodynamic,  237. 
Transverse  elasticity,  63. 
Truss,  Fink,  165. 

principal  rafter  of,  146. 

roof,  fixed  at  one  end,  free  at  the  other,  wind  and  dead  loads,  173. 
roof,  fixed  at  the  ends,  with  wind  pressure,  167. 
Trusses,  146. 

bays  or  panels  of,  146. 
braces  of,  146. 
chord  members  of,  146. 
web  members  of,  146. 
Tungsten  steel,  199. 
Twist,  angle  of,  96. 
Twisting  moment,  95. 

equivalent,  100. 

Ultimate  strength,  208. 


266  INDEX 

Vectors  of  force  diagram,  119. 
Velocity  of  flat  belts,  221. 
rope  belts,  234. 
Velocity  ratio,  244. 

in  flat  belt  transmission,  221. 

Warren  girder,  162. 
Web  of  I  beam,  68. 
Weight  of  cotton  belting,  220. 

leather  belting,  220. 

rope  belting,  234. 

rubber  belting,  220. 
Wind  pressure,  150. 

table  of,  150. 

Yield  point,  213. 
Zinc,  201. 


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